1. Introduction to Geometry and Topology

# 1. Cover both surfaces X and Y by patterns of polygons, and let uswrite V1 (resp. E1 resp. F1) for the number of vertices (resp. edges resp.faces) on X; let us write V2, E2, F2 for the corresponding numbers forthe surface Y . Let us assume that we chose these pattern such thatboth contain a face which is a polygon with k vertices and edges forsome k. The point of this is that removing that face from both surfacesand identifying the resulting surfaces along the polygon results in theconnected sum X#Y and the patterns on both surfaces fit together togive a pattern of polygons on X#Y . Let us determine V (= numberof vertices), E (= number of edges) and F (= number of faces) of thispattern on X#Y :

faces: F = F1 +F2−2 (each face on X or Y gives a face on X#Yexcept the two polygons that we removed in order to form theconnected sum).

edges: E = E1 + E2 − k (each edge on X or Y gives an edge onX#Y except there are k pairs of edges that get identified witheach other when we form the connected sum).

vertices: V = V1 + V2 − k (each vertex on X or Y gives a vertexon X#Y except there are k pairs of vertices that get identifiedwith each other when we form the connected sum).

This implies

χ(X#Y ) = V − E + F

= (V1 + V2 − k)− (E1 + E2 − k) + (F1 + F2 − 2)

= V1 − E1 + F1 + V2 − E2 + F2 − 2

= χ(X) + χ(Y )− 2

# 2. For g ≥ 0, let us write Σg for the surface of genus g. We notethat Σ0 = S2, Σ1 = T , and Σ2 = T#T and hence

χ(Σ0) = χ(S2) = 2 χ(Σ1) = χ(T ) = 0 χ(Σ2) = χ(T#T ) = −2

This suggests the formula

(1) χ(Σg) = 2(1− g)

for all g. Let us prove this statement by induction. To do the inductivestep, we assume that the above statement is true for all g ≤ N andneed to show that it is true for g = N + 1. We observe that

ΣN+1 = T# . . .#T︸ ︷︷ ︸N+1

= T# . . .#T︸ ︷︷ ︸N

#T = ΣN#T

1

2

Using the formula derived for the Euler characteristic of the connectedsum in the previous homework we obtain

χ(ΣN+1) = χ(ΣN#T ) = χ(ΣN) + χ(T )− 2

= 2(1−N) + 0− 2 = 2(1− (N + 1)),

i.e., the formula (1) holds for g = N + 1.

# 3. In order to calculate the Gaussian curvature of any surface Σ,we need to understand which curves on Σ are geodesics. In the case ofthe cylinder this can be done by thinking of the cylinder of radius ras being obtained by gluing the two sides of a strip of paper of infinitelength and width 2πr. Let us write C for the cylinder and S for thestrip of paper and f : S → C for the map that sends a point of the stripto the corresponding point on the cylinder. We note that for pointsp, q ∈ S which close to each other, the distance between p and q is thesame as that between their image points f(p), f(q) (as measured onthe cylinder). We note that this is not true for points which are furtherapart than πr.

Let us make this discussion more explicit. Let C ⊂ R3 be the cylinderconsisting of those points (x, y, z) ∈ R3 which have distance r from thex-axis and let S = {(x, y) ∈ R2 | 0 ≤ y ≤ 2πr}. Then we can describethe map f as

f : S −→ C (x, y) 7→ (x, r cos y/r, r sin y/r)

We note that the vertical line consisting of the points (0, y) maps to thecircle of radius r around the origin in the y − z−plane. In particularthe points p = (0, 0) and q = (0, 2πr) (which have distance 2πr inS) map to the same point in C; the points p = (0, 0), q = (0, 3

42πr)

have distance 342πr while their image points f(p) = (0, 1, 0), f(q) =

(0, 0,−1) have distance 142πr (their distance is measured in C, not

R3!).The fact that the map f preserves distances of point close to each

other implies that f maps geodesics in S to geodesics in C, and thatf preserves angles. In particular, f maps a geodesic triangle in Sto geodesic triangles in C with the same angle sum. Since geodesictriangles in S have angle excess 0, the same holds for geodesic trianglesin C. It follows that

K(p) = limT

e(T )

area(T )= 0

for all points p ∈ C.

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2. Chapter I

# 1.2.1.

(1) To show that the complement of a point x ∈ Rn is open, assumethat y ∈ Rn is in Rn − {x}; then the ball Oε(y,Rn) of radiusε := ||x − y|| does not contain the point x. Hence this ball iscontained in Rn − {x}; this shows that Rn − {x} is open.

(2) To show that the upper half plane is open, let y = (y1, y2) be apoint in the upper half plane. Then also the open ball Oε(y,R2)of radius ε := y2 is contained in the upper half plane. Hencethe upper half plane in open.

(3) To show that the rectangle R = (1, 2) × (5, 7) is open in R2,let y = (y1, y2) be a point of R. Then also the ball Oε(y,R2) ofradius ε := min{y1 − 1, 5− y1, y2 − 2, 7− y2} is contained in R(draw a picture!). To prove this, let x = (x1, x2) ∈ Oε(y,R2);then x1−y1 < ε ≤ 5−y1 and y1−x1 < ε ≤ y1−1 which implies1 < x1; similarly we get 2 < x2 < 7. Alternatively, this followsfrom Lemma 1.2.9(i).

# 1.2.7. Part (i) We need to show that if U ⊂ A, and V ⊂ B areopen subsets, then U ×V is an open subset of A×B. To show this, weneed to find for p = (p1, p2) ∈ U ×V an ε > 0 such that Oε(p,A×B) iscontained in U×V . Since U (resp. V ) is open in A (resp. B), we can findε1 > 0 (resp. ε2 > 0) such that Oε1(p1, A) ⊂ U (resp. Oε2(p2, B) ⊂ V ).This implies that the ‘rectangle’ Oε1(p1, A)×Oε2(p2, B) is contained inU × V (draw a picture for A = B = R!).Claim. For ε := min{ε1, ε2} the open ball Oε(p,A×B) is contained inthe rectangle Oε1(p1, A)×Oε2(p2, B).

This claim implies that U × V is open in A×B.While this claim is pretty clear from the picture for A = B = R, we

should provide a formal proof that works in the general case (for m > 1,n > 1, which we can’t visualize). We note that a point x = (x1, x2) isin the ball Oε(p,A×B if and only if ||x− p|| < ε; it is in the rectangleOε1(p1, A)×Oε2(p2, B) if and only if ||x1−p1|| < ε1 and ||x2−p2|| < ε2.So we assume ||x − p|| < ε and need to show ||x1 − p1|| < ε1 and||x2 − p2|| < ε2.

Let us abbreviate y = (y1, y2) = x− p = (x1 − p1, x2 − p2) ∈ Rn+m.Writing y out as y = (y1, y2, . . . yn+m), we have

y1 = (y1, . . . , yn) ∈ Rn and y2 = (yn+1, . . . , yn+m) ∈ Rm

4

and hence

||y||2 = (y1)2 + · · ·+ (yn+m)2

=((y1)2 + · · ·+ (yn)2

)+((yn+1)2 + · · ·+ (yn+m)2

)= ||y1||2 + ||y2||2.

Then

||x− p||2 = ||y||2 = ||y1||2 + ||y2||2 ≥ ||y1||2 = ||x1 − p1||2

and hence ||x1− p1|| ≤ ||x− p|| < ε ≤ ε1; the proof that ||x2− p2|| < ε2is completely analogous.

Part (ii) We need to show that every point p of an open set W ⊂A × B, there are numbers ε1 > 0, ε2 > 0 such that the ‘rectangle’Oε1(p1, A)×Oε2(p2, B) is contained in W .

To prove this, we note that by the assumption that W is open wecan find an ε > 0 such that the ball Oε(p,A × B) is contained in W .Then we would like to show that the rectangle Oε1(p1, A)×Oε2(p2, B)is contained in the ball Oε(p,A×B) for a suitable choice of ε1, ε2. Forma picture we see that the largest square fitting inside a ball of radius εhas sides of length ε/

√2. This suggests that the following claim should

be true. This claim implies what we want.Claim. For δ := ε√

2the ‘square’ Oδ(p1, A)×Oδ(p2, B) is contained in

the ball Oε(p,A×B).To prove the claim, let x = (x1, x2) be an element of Oδ(p1, A) ×

Oδ(p2, B); i.e., that ||x1 − p1|| < δ and ||x1 − p1|| < δ. We need toshow that x is in Oε(p,A × B); i.e., that ||x − p|| < ε. As in part (i)we write y = x − p and hence using the norm calculations from part(i) we obtain

||x−p||2 = ||y||2 = ||y1||2 + ||y2||2 = ||x1−p1||2 + ||x1−p1||2 < 2δ2 = ε2

Taking square roots we conclude ||x− p|| < ε, which proves the claim.

# 1.2.8. To show that the following subsets are closed in R2, weconsider their complements and use Lemma 1.2.9 according to whichthe cartesian product of open subsets are open.

(1) The complement of R× {0} ⊂ R2 is union of R× (−∞, 0) andR× (0,∞). Both of these sets are open by Lemma 1.2.9. Hencetheir union is open and R× {0} is closed.

(2) The complement of the upper half plane H2 is the open lowerhalf plane R × (−∞, 0), which is open by Lemma 1.2.9 andhence H2 is closed.

(3) The complement of [1, 2]× [5, 7] is the union of the sets (∞, 1)×R, (5,∞)× R, R× (−∞, 2) and R× (7,∞) (draw a picture!).

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These are open sets by Lemma 1.2.9, and hence their union isopen. This shows that [1, 2]× [5, 7] is closed.

# 1.2.15. To prove that the closed ball Or(p,A) is a closed subset ofA, we need to show that the complement A−Or(p,A) is an open subsetof A. So let x be a point in the complement, i.e., ||x − p|| > r. Weclaim that then the ball Oε(x,A) of radius ε := ||x−p||−r is containedin the complement A − Or(p,A) (draw the picture). To prove this,assume y ∈ Oε(x,A). Then by the triangle inequality we have

||x− p|| ≤ ||x− y||+ ||y − p|| < ε+ ||y − p|| = ||x− p|| − r + ||y − p||,

which implies ||y − p|| > r; in other words, y is in the complementA− Or(p,A), which proves our claim.

This proves that A− Or(p,A), i.e., Or(p,A) is a closed subset of A.

# 1.2.16. To show that U − {x} is open in A, consider a pointy ∈ U − {x}. Since U is open in A, there is some r > 0 such that theball Or(y, A) is contained in U . Then the possibly smaller ball Oε(y, A)with ε := min{r, ||x− y||} is also contained in U . In fact, the ε-ball iscontained in U − {x}, since x /∈ Oε(y, A). This shows that U − {x} isopen in A.

# 1.2.17. To show that the least upper bound s of a closed subsetS ⊂ R is contained in S, we use proof by contradiction. Let us assumes 6= S; i.e., s is an element of the complement R− S. This is an openset by assumption and hence we find some open ball Oε(s,R) that iscontained in R − S. This is the desired contradiction since it impliesthat s− ε

2is still an upper bound for S, contradicting our assumption

that s is the least upper bound. The argument for the greatest lowerbound is analogous.

#1.3.1 To show that the inclusion map i : B → A is continuous, letU ⊂ A be an open set. Then

i−1(U) = {b ∈ B | i(b) ∈ U} = {b ∈ B | b ∈ U} = B ∩ U

We emphasize that B is not assumed to be open, so B ∩ U is notnecessarily an open subset of A. However, to show continuity of i, weneed to show that i−1(U) is an open subset of the domain of i, whichis B. This follows from Lemma 1.2.8.

# 1.3.3 We note that the map f|B : B → C (the restriction of f toB ⊂ A) is equal to the composition of the inclusion map i : B → Awith the map f : A → C. By the previous problem, i is continuous,and hence so is f ◦ i as the composition of two continuous maps.

6

# 1.3.5 We want to show that a bijective map f : A → B is open ifand only if it is closed. First we prove the following statement:Claim. If F : A→ B is bijective, then f(A− U) = B − f(U).

To prove this, assume y ∈ f(A − U); i.e., y = f(x) with x ∈ A,x /∈ U . We want to show y ∈ B−f(U); i.e., that y is not an element off(U). Using proof by contradiction, we assume y ∈ f(U), i.e., y = f(u)for some u ∈ U . Then f(u) = y = f(x) implies u = x (by injectivityof f). This is the desired contradiction, since u ∈ U , but x /∈ U .

Conversely, assume y ∈ B − f(U); i.e., y /∈ f(U). We want to showy ∈ f(A− U); i.e., y = f(x) for some x ∈ A− U . By surjectivity of f ,there is some x ∈ A with y = f(x). The element x can’t be in U byour assumption y /∈ f(U) and hence x ∈ A− U as desired.

Now we want to show:

f is open =⇒ f is closed.

Let C be a closed subset of A. Then A−C is open and hence f(A−C) isopen (since f is assumed open). Hence B−f(C) = f(A−C) (equalityby the claim above) is open, which implies f(C) is closed. This showsthat f is a closed map.

Conversely, we want to show

f is cosed =⇒ f is open.

This follows the same argument as above, replacing ‘open sets’ by‘closed sets’ and vice versa: Let U ⊂ A be open. Then A − U isclosed and hence f(A − U) is closed (since f is assumed closed). Butf(A−U) = B−f(U) (by claim) and hence f(U) is open, which finishesthe proof that f is open.

# 1.3.8 To show that the function h := max{f, g} is continuous, wewill use Lemma 1.3.6. We note that

h|A1 = f|A1 for A1 = {x ∈ R | f(x) ≥ g(x)}

and

h|A2 = g|A2 for A2 = {x ∈ R | g(x) ≥ f(x)}and A1 ∪ A2 = R. Moreover, f , g are continuous by assumption, andso the restrictions f|A1 , g|A2 are continuous (by a previous homeworkproblem). So Lemma 1.3.6 lets us conclude that h is continuous, pro-vided we can show that both subsets A1, A2 are either open or closed.Drawing a picture suggests that they should be closed. To prove this,

7

write A1 in the following form:

A1 = {x ∈ R | f(x) ≥ g(x)} = {x ∈ R | f(x)− g(x) ≥ 0}= {x ∈ R | k(x) ∈ [0,∞)} = k−1([0,∞)),

where k : R→ R is defined by k(x) = f(x)−g(x). Appealing to a resultproved in calculus, the continuity of f and g implies the continuity ofk. Hence the preimage k−1([0,∞)) of the closed subset [0,∞) ⊂ R isclosed. Also A2 = k−1((−∞, 0]) is closed.

A similar argument shows that the function min{f, g} is continuous.

Additional problem # 1Let U be the standard topology on Rm+n; i.e., W ∈ U if and only ifW

is open in the sense that for every point p ∈ W there is a r > 0 such thatOr(p,Rm+n) ⊂ W . Let P be the product topology on Rm×Rn = Rm+n

consisting of those subsets W which are unions of subsets of the formU × V , where U is an open subset of Rm, and V is an open subset ofRm.

To show that U = P , assume W ∈ U , i.e., W is an open subsetof Rm+n. Then according to part (ii) of Lemma 1.2.9 for each pointp = (p1, p2) ∈ W there exist r1(p), r2(p) > 0 such that the productOr1(p)(p1,Rm)×Or2(p)(p2,Rn) is contained in W . In particular,

W =⋃p∈W

Or1(p)(p1,Rm)×Or2(p)(p2,Rn),

which shows W ∈ P .Conversely, if W ∈ P , then W =

⋃i∈I Ui × Vi with Ui open in Rm

and Vi open in Rn. By part (i) of Lemma 1.2.9, Ui×Vi is open in Rm+n,i.e., Ui × Vi ∈ U . Hence W as the union of these sets in also in U .

Additional problem # 2 Let (X,U), (Y,V) be topological spaces,and let A ⊂ X, B ⊂ Y be subsets. We want to show that the producttopology P on A×B (where A and B are equipped with their subspacetopologies) agrees with the subspace topology S (given by regardingA×B as a subset of X × Y equipped with the product topology).

Suppose that W ∈ P . Then W is a union of subsets of the form U×Vwith U belonging to the subspace topology of A and V belonging tothe subspace topology of B. By the definition of the subspace topologythis means that are subsets U ′ ∈ U , V ′ ∈ V such that U = U ′ ∩A andV = V ′ ∩B and hence

U × V = (U ′ ∩ A)× (V ′ ∩B) = (U ′ × V ′) ∩ (A×B).

8

We note that U ′ × V ′ belongs to the product topology on X × Y , andhence U ×V S. Then also W ∈ S, since W is a union of sets belongingto S.

Conversely, if W ∈ S, then W is of the form W = W ′ ∩ (A × B),where W ′ ⊂ X × Y belongs to the product topology on X × Y . Thismeans that W ′ can be written in the form

W ′ =⋃i∈I

U ′i × V ′i

with U ′i ∈ U and V ′i ∈ V . Then

W = W ′ ∩ (A×B) =⋃i∈I

(U ′i × V ′i ) ∩ A×B =⋃i∈I

(U ′i ∩ A)× (V ′i ∩B).

We note that U ′i ∩A belongs to the subspace topology of A and Vi ∩Bbelongs to the subspace topology of B which implies W ∈ P .

# 1.4.2. We want to show that a map f : A→ B is a homeomorphismif and only if it satisfies the following two conditions:

(1) f is bijective(2) for every subset U ⊂ B the set U is open if and only if its

pre-image f−1(U) is open in A.

We recall that a map f : A → B is a homeomorphism if it is bijectiveand if f as well as its inverse map g : B → A are continuous (we writeg instead of the usual f−1 here to avoid confusion when dealing withpre-images under g).

First we want to prove that if f is a homeomorphism, then it satisfiesconditions 1. and 2. above. By definition, condition 1. is satisfied, andthe continuity of f implies that if U is open in B, then f−1(U) is openin A. So will still need to show that if f−1(U) is open in A, then U isopen in B. To prove this, we note that U = g−1(f−1(U) (do a proofof that!), and hence the continuity of g implies that if f−1(U) is open,then so is U .

Conversely, now we want to assume that f satisfies conditions 1. and2., and we want to argue that f is a homeomorphism. The only thingleft to show is that the inverse map g is continuous. So let us assumethat V is open in A (the range of g) and we want to prove that g−1(V )is open in B (the domain of g). Let us set U := f(V ) ⊂ B; then V =f−1(U) (convince yourself of that) and g−1(V ) = g−1(f−1(U)) = U .Hence condition 2. implies that the openness of V = f−1(U) impliesthat g−1(V ) = U is open. This shows that g is continuous.

9

# 1.4.1

(1) To show that any open disk in R2 is homeomorphic to R2, wefirst note that any open disk Or(y,R2) is homeomorphic to thedisk Oπ/2(0,R2): the map

g : Or(y,R2) −→ Oπ/2(0,R2) x 7→ π/2

r(x− y)

is a continuous map (since its components are continuous) whoseinverse is the following map (which is continuous by the sameargument):

g−1 : Oπ/2(0,R2) −→ Or(y,R2) x 7→ r

π/2x+ y

So it now suffices to produce a homeomorphism between thedisc Oπ/2(0,R2) and R2. The map

f : Oπ/2(0,R2) −→ R2 x 7→ tan ||x||||x||

x

is a continuous map, since its components are continuous (notethat tan t

textends to a continuous map with domain [0,∞)). Its

inverse is given by

f−1 : R2 −→ Oπ/2(0,R2) x 7→ arctan ||x||||x||

x,

which is continuous by the same argument (note that this iswell-defined since arctan t

textends to a continuous map with do-

main [0,∞)).(2) We’ve shown in class that any interval [a, b] is homeomorphic

to R; in particular, there are homeomorphisms

f : [a, b] −→ R and g : [c, d] −→ R.Then f × g : [a, b] × [c, d] → R × R = R2 given by (x, y) 7→(f(x), g(x)) is a continuous map, since its components are con-tinuous. By the same argument,

f−1 × g−1 : R2 → [a, b]× [c, d]

is continuous. This is the inverse map to f × g, and so f × g isa homeomorphism.

# 1.4.5 Let q : X → Y be a quotient map. We want to show that amap f : Y → Z is continuous if and only if the composition f ◦g : X →Z is continuous.

10

One implication is straightforward: if f is continuous, then so is f ◦gas the composition of two continuous maps. Conversely, let us assumethat f ◦ q is continuous. In order to show that f is continuous, weassume that U ⊂ Z is open, and need to argue that g−1(U) is an opensubset of Y . Continuity of f ◦q implies that (f ◦q)−1(U) = q−1(f−1(U))is open in X. Since q is a quotient map, this implies that f−1(U) isopen in Y which we had to show.

# 1.4.6 In this problem, I will describe the identification space Y andwill give a map q from X to Y , but I won’t prove that q is a quotientmap and that the pre-images q−1(y) agree with the given partition (seethe following problem for what I expect along these lines...).

(1) We show that the identification space is the 2-sphere S2 = {x ∈R3 | ||x|| = 1} using the quotient map

q : O1(0,R2) −→ S2

x = (x1, x2) 7→ (cos π||x||, sin π||x||||x||

x1,sin π||x||||x||

x2);

we note that the above map is well-defined at x = 0, sincelimt→0

sinπtt

= π, which shows that this function can be ex-tended to a continuous function with domain [0,∞).

(2) We show that the identification space is R2 using the quotientmap

q : R2 −→ R2 given by q(x) =

{0 ||x|| ≤ 1

(||x|| − 1) x||x|| ||x|| ≥ 1

(3) We show that the identification space is the circle S1 = {x ∈R2 | ||x|| = 1} using the quotient map

q : R −→ S1 given by q(t) = (cos 2πt, sin 2πt)

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# 1.4.7

(1) We claim that the attaching space X ∪h Y is [0, 3]. To provethis, we need to find a quotient map

q : [0, 2] ∪ [3, 5] −→ [0, 3] with {q−1(z) | z ∈ [0, 3]} = P(h).

Let us define

q(t) =

{t t ∈ [0, 2]

t− 2 t ∈ [0, 3]

We note that• q is surjective (since the image of q restricted to [0, 2] is

[0, 2], while the image of q restricted to [3, 5] is [1, 3], andhence the image of q is [0, 2] ∪ [1, 3] = [0, 3]);• q is continuous, since its restriction to [0, 2] (resp. [3, 5])

is continuous, and these are both open (as well as closed)subsets of the domain of q;• We note that

q−1(z) =

{z} z ∈ [0, 1)

{z, z + 2} z ∈ [1, 2]

{z + 2} z ∈ (2, 3]

,

showing that the preimages q−1(z) agree with the partitionP.

This shows that q is a quotient map (skipping as usual here theproof that q is open) and hence [0, 3] is the attaching space.

(2) We claim that the attaching space X ∪h Y is homeomorphic tothe cone

C = {(x, y, z) ∈ R3 | ||(x, y)|| < 1, z = ±||(x, y)||} ⊂ R3.

To prove this claim consider the map

q : X ∪ Y = O1((1, 0),R2) ∪O1((4, 0),R2) −→ C

defined by

q(x, y) =

{(x− 1, y, ||(x− 1, y)||) (x, y) ∈ O1((1, 0),R2)

(x− 4, y,−||(x− 4, y)||) (x, y) ∈ O1((4, 0),R2)

We note that• The restriction of q to the disjoint open setsX = O1((1, 0),R2)

(resp. Y = O1((4, 0),R2)) is continuous, since all compo-nents are continuous functions. Hence q is continuous.

12

• The image of the restriction q|X is the upper half-cone C ∩{(x, y, z)R3 | z ≥ 0}, the image of q|Y is the lower half-coneand hence q is surjective onto the cone C.• The restriction of q to either X or Y is injective; hence the

only points that map to the same point under q are thecenters (1, 0) resp. (4, 0) of the two discs that both map to~0 = (0, 0, 0) ∈ R3. Hence the preimages q−1(z) consist of

all of single points except for q−1(~0) = {(1, 0), (4, 0)}; inparticular, these preimages agree with the partition P(h)of X ∪ Y .

This shows (except for the openness of q) that the cone C isthe attaching space X ∪h Y .

# 1.5.4 We need to show:

(i) The components of a space A are disjoint;(ii) Each component is closed;(iii) A is the union of all components

If A has finitely many components, say C1, . . . , Ck, then each compo-nent is also open since the complement A − Cj is equal to the union⋃i 6=j Ui of finitely many closed set and hence closed. If A has infin-

itely many components, these components are not necessarily open.For example, the space A = {0, 1, 1

2, 1

3, . . . , 1

n, . . . } has infinitely many

components, namely each point of this space constitutes a component(prove this!). We note that the component {0} is not open in A, sinceany ball around 0 contains some other points of A.Proof of (ii). Let C ⊂ A be a component of A, i.e.,

(1) C is connected(2) C is non-empty(3) C is not a proper subset of a connected subset of A

We need to show that C is closed, i.e., that A\C is open. We argue asfollows: for x ∈ A \C, consider C ∪ {x}, which contains C as a propersubset. Hence by (3) above, C ∪{x} is not connected, and hence it canbe decomposed in the form

C ∪ {x} = B1 ∪B2,

where B1, B2 are disjoint open non-empty subsets of C ∪{x}. W.l.o.g.,x ∈ B1. Then

C = (C ∩B1) ∪ (C ∩B2),

13

and C ∩ B1 C ∩ B2 are open disjoint subsets of C. Since x /∈ B2,we see that C ∩ B2 is equal to B2; in particular, it is non-zero. Thisimplies that C∩B1 must be empty (otherwise the above decompositionof C would be a decomposition into disjoint non-empty open subsetscontradicting the assumption that C is connected).

We conclude B1 consists only of the point x. By construction B1

is open in C ∪ {x} (with respect to the subspace topology) and hencethere is some open subset Ux ⊂ A such that Ux∩(C∪{x}) = B1 = {x};in particular, Ux ∩ U = ∅. If follows that the union U =

⋃x∈A\C Ux is

an open subset of A; this subset is equal to A \ C and hence A \ C isopen in A, which is what we need to show for (2).Proof of (i). We need to show that if C1 and C2 are components whichare not disjoint, then C1 = C2. So let x ∈ C1 ∩ C2. We claim thatC1 ∪ C2 is again connected. We use proof by contradiction: assumeC1 ∪ C2 is not connected; then it can be decomposed as a union

C1 ∪ C2 = B1 ∪B2

of non-empty, disjoint open subsets of C1∪C2. Assume w.l.o.g. x ∈ B1,and C1 ∩B2 6= ∅, and consider the decomposition

C1 = (C1 ∩B1) ∪ (C1 ∩B2).

We note that C1 ∩ B1 and C2 ∩ B2 are open disjoint and non-emptysubsets of C1 (C1 ∩B1 is non-empty since it contains x). This contra-dicts the fact that C1 as a component of A is connected, thus provingour claim that C1 ∪ C2 is connected. By property (3) of a component,this implies that C1 can’t be a proper subset of C1∪C2, i.e., C1 must beequal to C1 ∪C2. Similarly we have C2 = C1 ∪C2, and hence C1 = C2

as desired.Proof of (iii). We need to show that any point x ∈ A is contained insome component C of A. To prove this, let C be the union of all con-nected subsets of A containing x. By problem 1.5.2 (or by generalizingour proof above that C1∪C2 is connected) this union is connected; it isnon-empty since it contains x; it is not a proper subset of a connectedsubset, since any such set D is one of the sets in the union that definesC. This shows that C is a component; clearly it contains x.

# 1.5.6 We want to prove the Brower Fixed Point Theorem in di-mension one; i.e., we want to show that if f : [a, b] → [a, b] is a con-tinuous map, then f has a fixed point, i.e., a point d ∈ [a, b] such thatf(d) = d. The idea is to note that d ∈ [a, b] is a fixed point if and only

14

if f(d) − d = 0; this suggests to apply the Intermediate Value to thecontinuous function g(x) := f(x) − x and to look for points d ∈ [a, b]with g(d) = 0. We note that g(a) = f(a) − a ≥ 0 (since f(a) ∈ [a, b])and g(b) = f(b) − b ≤ 0 (since f(b) ∈ [a, b]). Hence the Intermedi-ate Value Theorem guarantees the existence of some d ∈ [a, b] withg(d) = 0 and so f(d) = d as desired.

# 1.5.7

(1) To show that any open ball Or(p,Rn) in Rn is path connected,let x, y ∈ Or(p,Rn), and consider the linear path

c : [0, 1] −→ Or(p,Rn) c(t) = tx+ (1− t)y.

This path is continuous since its components ci(t) = txi + (1−t)yi are continuous functions (here x = (x1, x2, . . . , xn) and y =(y1, y2, . . . , yn)). It remains to show that this is in fact a pathin Or(p,Rn) (geometrically this is pretty obvious); i.e., we needto show that ||c(t) − p|| < r for all t ∈ [0, 1]. The followingcalculation proves this:

||c(t)− p|| = ||tx+ (1− t)y − p|| = ||tx+ (1− t)y − tp− (1− t)p||= ||t(x− p) + (1− t)(y − p)||≤ ||t(x− p)||+ ||(1− t)(y − p)|| = t||x− p||+ (1− t)||y − p||< tr + (1− t)r = r

Here the first inequality is a consequence of the triangle inequal-ity; the second comes from ||x−p|| < r and ||x−p|| < r, whichhold since x, y are by assumption in the open ball Or(p,Rn).

(2) To show that the ‘punctured open ball’ X = Or(p,Rn) − {p},is path connected, let x, y ∈ X; if y is not on the line throughx and p, the path c(t) constructed in part (1) is a path in X.If x, y and p are all on one line, choose a point z not on thisline (this is possible provided n ≥ 2); then use the constructionin (1) to produce a path c1 from x to y (i.e., c1 is a continuousmap c1 : [0, 1] → X with c1(0) = x and c1(1) = z) as well asa path c2 from z to y. Then we ‘concatenate’ these paths toproduce a new path c from x to y; explicitly, c : [0, 1] → X isgiven by

c(t) :=

{c1(2t) 0 ≤ t ≤ 1

2

c2(2t− 1) 12≤ t ≤ 1

15

We note that c is well-defined, since for t = 12

the values c1(2t) =c1(1) = z and c2(2t − 1) = c2(0) = z agree; moreover, c iscontinuous, since its restriction to [0, 1

2] resp. [1

2, 1]

(3) To show that the unit circle S1 is path connected, we note thatthe map f : R→ S1 given by f(t) = (cos t, sin t) is continuous,since its components are continuous. By Problem 1.5.9 the factthat R, the domain of f , is path connected implies that theimage f(R) is path connected. Since f is surjective, its imageis all of S1, and so S1 is connected.

# 1.5.9 Let f : A → B be a continuous map and assume that A ispath connected. To prove that the image f(A) is path connected, weneed to show that any two points x, y ∈ f(A) can be connected by apath. Since x, y are in the image of f , there are elements x′, y′ ∈ Awith f(x′) = x and f(y′) = y. Since A is path connected, there is apath in A connecting x and y, i.e., a continuous map c : [0, 1]→ A withc(0) = x′ and c(1) = y′. Then the composition c′ := f ◦ c : [0, 1] → Bis a path in B connecting c′(0) = f(c(0)) = f(x′) = x and c′(1) =f(c(1)) = f(y′) = y which is what we needed to show.

Extra problem. We want to prove that the cartesian product X =X1×· · ·×Xn of path connected spacesX1, . . . , Xn is path connected. Sowe need to show that for any points x, y ∈ X, there is a path connectingthese points, i.e., a continuous map c : [0, 1] → X with c(0) = x andc(1) = y. Let us write

x = (x1, . . . , xn), y = (y1, . . . yn) with x1, y1 ∈ X1, . . . , xn, yn ∈ Xn.

SinceXi is path connected there is a path ci : [0, 1]→ Xi with ci(0) = xiand ci(1) = yi. Now define c : [0, 1]→ X by

c(t) := (c1(t), . . . , cn(t)) ∈ X1 × · · · ×Xn = X

This is a continuous map since its components are continuous and wehave

c(0) = (c1(0), . . . , cn(0)) = (x1, . . . , xn) = x

c(1) = (c1(1), . . . , cn(1)) = (y1, . . . , yn) = y.

This shows that X is path connected.

16

# 1.6.1 Let A1, . . . , Ak be compact subsets of Rn. To show that A :=A1 ∪ · · · ∪ Ak is compact, let {Ui}i∈I be an open cover of A. Then{Ui ∩Aj} is an open cover of Aj. By compactness of Aj, this containsa finite subcover; i.e., there is a finite subset Ij ⊂ I such that Aj iscontained in the union

⋃i∈Ij Ui. This implies that A is contained in

the union of the Ui for i in the finite set J := I1 ∪ · · · ∪ Ik; in otherwords, {Ui}i∈J is a finite subcover of {Ui}i∈I .

# 1.6.4 We want to find spaces A, B and a continuous bijectionf : A→ B such that f is not a homeomorphism. The hint in the booksuggests to take A = R and to look for an injective map f : A → R2

which is not a homeomorphism onto its image B := f(A). We note thatgeometrically, the map f describes a path in R2 which is parametrizedby R. I find it easier to come up with such paths parametrized by ahalf-open interval, say A = [0, 1). Consider the path f : [0, 1) → R2

given by f(t) = (cos 2πt, sin 2πt). This is a continuous bijection, but itis not a homeomorphism, since f is not an open map. To see this, wenote that [0, 1

2) is an open subset of [0, 1) whose image consists of the

upper semi-circle {(x, y) | x2 + y2 = 1, y > 0} and the point p = (1, 0).This is not an open subset of S1, since any open ball Oε(p, S

1) containsalso points in the lower half circle.

Later I noticed that the above example does not satisfy the require-ment that the domain be a closed subset. We can fix this by choosingA′ = [0,∞), and f ′ = g ◦ f : A′ → R2, where g : [0,∞) → [0, 1) is thehomeomorphism given by g(x) = arctanx/(π/2). Then f ′ is a con-tinuous bijection with closed domain which is not a homeomorphism(otherwise, f = g−1 ◦ f ′ would be a homeomorphism).

This example also shows that a surjective continuous map f : A →B is not necessarily a quotient map (the map construct above is asurjective continuous map, but not a quotient map, since it is notopen.

# 1.6.6 To show that a closed interval is not homeomorphic to R, wenote that a closed interval is compact, but R is not; hence by problem1.6.3, these spaces cannot be homeomorphic.

17

# 1.6.8 We want to show that a compact subset A ⊂ R has a minimaland a maximal member; i.e., there are points x1, x2 ∈ A such thatx1 ≤ x ≤ x2 for all x ∈ A. By the Heine-Borel Theorem, A is closedand bounded. In particular, A has upper bounds as well as lowerbounds, and we can define

x1 := glbA x2 := lubA.

Since S is closed, x1, x2 belong to A by problem 1.2.17.

# 1.6.11 We want to show that for disjoint open sets A,B ⊂ Rn thereis a number m > 0 such that ||a − b|| ≥ m for all a ∈ A, b ∈ B.Consider the map

f : A×B −→ R (a, b) 7→ ||a− b||.

To see that f is continuous, note that it is the restriction of the com-position

Rn × Rn g−→ Rn h−→ R,to A×B ⊂ Rn×Rn, where g(a, b) = a− b and h(x) = ||x||. The mapsg, h are both continuous, and hence so is f .

We note that g is continuous since all its component functions arecontinuous (if we write a = (a1, . . . , an) and b = (b1, . . . , bn), then thei-th component function gi(a, b) is equal to ai− bi). To argue that h iscontinuous, we use the ε − δ-characterization of continuity and claimthat for given ε > 0, we can find δ > 0 such that

||x− y|| < δ =⇒ |h(x)− h(y)| < ε.

To see this, we note

||x|| = ||y+(x−y)|| ≤ ||y||+||x−y|| and hence ||x||−||y|| ≤ ||x−y||.

Similarly,

||y|| = ||x+(y−x)|| ≤ ||x||+|y−x|| and hence ||x−y|| ≤ ||x||−||y||

Putting these two inequalities together we see |h(x) − h(y)| = |||x|| −||y||| ≤ ||x − y|| which shows that we can choose δ = ε above; inparticular, h is continuous.

According to Proposition 1.6.12, the function f has a minimum valueon its compact domain A × B, i.e., there exist some (a0, b0) ∈ A × Bsuch that for any (a, b) ∈ A × B we have f(a, b) ≥ f(a0, b0). We setm := f(a0, b0) and note that m > 0 (if f(a0, b0) = ||a0 − b0|| = 0,

18

then a0 = b0 which contradicts the assumption that A, B are disjoint).Then

||a− b|| = f(a, b) ≥ f(a0, b0) = m ∀a ∈ A, b ∈ B.

# 2.2.3 Let J ⊂ Rn be an arc, and let h1, h2 : [0, 1]→ J be homeomor-phisms. We need to show that h1((0, 1)) = h2((0, 1)) and h1({0, 1}) =h2({0, 1}). This statement is the 1-dimensional analog of Lemma 2.2.3(which makes the analogous assertion for disks), and our proof willbe modeled on the proof of that lemma. Let f : [0, 1] → [0, 1] be thehomeomorphism given by the composition

[0, 1]h1 //J

h−12 // [0, 1].

Then it suffices to show that f((0, 1)) = (0, 1) and f({0, 1}) = {0, 1}(this is equivalent to the equations h1((0, 1)) = h2((0, 1)) and h1({0, 1}) =h2({0, 1}) as can be seen by applying h2 resp. h−1

2 to both of them).By a previous homework problem, we know that (0, 1) is homeomor-

phic to R. This shows that f((0, 1)) ⊂ [0, 1] ⊂ R is homeomorphic toR and hence by the Invariance of Domain Theorem f((0, 1)) is an opensubset of R. Hence 0 and 1 cannot be contained in f((0, 1)), since anyball around those points contain points outside of [0, 1], which cannotbe in the image f((0, 1)). It follows that f((0, 1)) ⊂ (0, 1). Applyingthe same argument to f−1 : [0, 1]→ [0, 1] shows f−1((0, 1)) ⊂ (0, 1) andhence (0, 1) = f(f−1((0, 1))) ⊂ f((0, 1)). Putting these two inclusionstogether we conclude f((0, 1)) = (0, 1).

To prove f({0, 1}) = {0, 1}, we note that {0, 1} is the complementof (0, 1) ⊂ [0, 1]; hence the image of {0, 1} under the bijection f mustbe the complement of f((0, 1)) ⊂ [0, 1], which is {0, 1}.

# 2.2.5 Let C ⊂ R2 be a 1-sphere. We need to show:

(i) (Jordan Curve Theorem) The set R2−C has precisely two com-ponents, one of which is bounded and the other of which isunbounded.

(ii) The union of C and the bounded component of R2−C is a diskof which C is the boundary.

To prove the Jordan Curve Theorem, we argue that by the SchonfliesTheorem, there is a homeomorphism H : R2 → R2 such that H(S1) =C and H is the identity map outside a disk. Restricting H to the

19

complement R2 − S1 then gives a homeomorphism

R2 − S1 ≈−→ R2 − C.We recall that if two spaces are homeomorphic, and one is connected,then so is the other (by Theorem 1.5.3). Applying this to subspacesof a space X, we see that if A ⊂ X is a component of X, and iff : X → Y is a homeomorphism, then f(A) ⊂ Y is a component ofY . In particular, we see that R2 − C has two components, namelyH(intD2) and H(R2 −D2).

We claim that H(D2) and hence H(intD2) is bounded. Note that‘boundedness’ is not a property preserved by homeomorphisms (e.g.,the bounded subset (0, 1) ⊂ R is homeomorphic to all of R); conse-quently, we can’t just argue that since intD2 is bounded, then so is itimage under the homeomorphism H. To prove that H(D2) is bounded,we use that fact that H is the identity outside of the ball OR(0,R2)for some R > 1. This implies that H maps OR(0,R2) to itself; in par-ticular, it maps D2 ⊂ OR(0,R2) to a subset of OR(0,R2), and henceH(D2) is bounded.

We claim that H(R2−D2) is unbounded. We use proof by contradic-tion: we assume that H(R2−D2). Then H(R2) = H(D2)∪H(R2−D2)is the union of two bounded subsets of and hence bounded. This impliesthat H can’t be surjective, which is the desired contradiction.

To show that the union of C and the bounded component H(intD2)is a disk is now straightforward. Let us write B := C ∪ H(intD2).Then

B = C ∪H(intD2) = H(S1) ∪H(intD2) = H(S1 ∪ intD2) = H(D2),

and hence H|D2 : D2 → B is a homeomorphism. This implies that B isa disk with boundary H(S1) = C.

# 2.2.13 We will use proof by contradiction to show that no propersubset of S1 is homeomorphic to S1. So assume that A ⊂ S1 is a propersubset of S1 which is homeomorphic to S1. Then A is a 1-sphere in R2

and hence by the Jordan Curve Theorem, the complement R2 \ A hastwo connected components. We will show that R2\A is path connectedhence obtaining the desired contradiction.

We decompose the complement R2 \ A as follows:

R2 \ A = (R2 \D2) ∪ (D2 \ A).

We note that D2 \A is path connected since any point v ∈ D2 \A canbe connected with 0 ∈ R2 via the radial path c : [0, 1] → D2 \ A given

20

by c(t) = tv (and hence by concatenating paths, any point in D2 \ Acan be connected by any other point). We claim that also R2 \ D2

is path connected. One way to proving this is to note that there is ahomeomorphism

R2 \ {0} −→ R2 \D2 x 7→ ||x||+ 1

||x||x

(the inverse is given by x 7→ ||x||−1||x|| x). In homework problem 1.5.7

we have shown that R2 \ {0} is path connected. It follows that thehomeomorphic space R2 \D2 is path connected since ‘path connected’in an intrinsic property).

By assumptionA is a proper subset of S1 and hence there is some unitvector v ∈ R2 which doesn’t belong to A. Then the path c : [0, 1]→ R2,t 7→ 2tv is in the complement of R2 \ A connecting the point c(0) =0 ∈ D2 \ A and c(1) = 2v ∈ R2 \ D2. Again by concatenating paths,this shows that any two points in R2 \ A can be connected by a path.Hence R2 \ A is path connected and hence connected.

# 2.3.1 No sample solution provided

# 2.3.5 There is a frequently used homeomorphism

h : S2 − {n} −→ R2,

called the stereographic projection (where n = (0, 0, 1) ∈ S2 is thenorth pole). For x ∈ S2 − {n}, the point h(x) ∈ R2 can be describedgeometrically as follows: the line through the north pole n and thegiven point x intersects the plane R2 ⊂ R3 in exactly one point; thisis the point h(x) ∈ R2. The map h is a bijection; the inverse maph−1 : R2 → S2 − {n} is given by sending a point y ∈ R2 to the pointwhere the line through n and y intersects the sphere S2. To show thath and h−1 are continuous, we use elementary geometry to write h andh−1 explicitly as functions and we will see that the coordinate functionsare in fact continuous.

We observe that if p is any point of S2, there is an orthogonal lineartransformation A : R3 → R3 mapping p to n (find two orthonormalbases {p1, p2, p3} and {n1, n2, n3} such that p1 = p and n1 = n; thenthere is a unique orthonormal linear transformation A with A(pi) = nifor i = 1, 2, 3). We note that A (as a orthogonal transformation) mapsunit vectors to unit vectors and so gives a continuous map A : S2 → S2;

21

its inverse A−1 : S2 → S2 is again a continuous map and hence A is ahomeomorphism. The composition

g : S2 − {p} A−→ S2 − {n} h−→ R2

is then the desired homeomorphism which proves part (i).To prove part (ii), we restrict g to the complement of q ∈ S2 − {p}

to obtain a homeomorphism

g : S2 − {p, q} −→ R2 − {z} z = g(q).

Let T : R2 → R2 be the ‘translation’ homeomorphism given by T (x) =x − y (its inverse is given by adding y). Restricting T results in thehomeomorphism T : R2 − {y} ∼= R2 − {0}. Finally, let

f : R2 − {0} −→ S1 × R be given by f(x) = (x

||x||, ln ||x||).

This is a continuous map (since its components are continuous) andit is in fact a homeomorphism since its inverse is the continuous mapS1 × R→ R2 − {0} given by (y, r) 7→ ery. Then the composition

S2 − {p, q} g−→ R2 − {z} T−→ R2 − {0} f−→ S1 × Ris the desired homeomorphism of part (ii).

# 2.3.7 Let H1 : Q1− intB1 → Q2− intB2 be a homeomorphism, andlet h : ∂B1 → ∂B2 be its restriction to the boundary.

Claim. h extends to a homeomorphism H2 : B1 → B2.Assuming the claim for now, let H : Q1 → Q2 be the map defined by

H(x) =

{H1(x) x ∈ Q1 − intB1

H2(x) x ∈ B1

We note thatH is well defined sinceH1 andH2 agree on the intersectionof Q1−intB1 and B1 by construction. Since H1, H2 are continuous andB1 and Q1− intB1 are closed subsets of Q1, the map H is continuous.Reversing the roles of Q1 and Q2 we can use the same construction toproduce a continuous map Q2 → Q1 which inverse to H; this showsthat H is a homeomorphism.

To prove the claim, let hi : D2 ≈−→ Bi be homeomorphisms defining

the disks Bi for i = 1, 2. Then the restriction (hi)|S1 of hi to S1 ⊂ D2

is a homeomorphism from S1 to ∂Bi. Let f be the composition

f : S1(h1)|S1

−→ ∂B1h−→ ∂B2

(h2)−1

|S1

−→ S1

22

and let F : D2 → D2 be the ‘radial extension’ of f to the disk definedby

F (x) =

{||x||f( x

||x||) x 6= 0

0 x = 0.

It is clear that F is an extension of f (i.e., that the restriction of F toS1 is f) and that F restricted to D2\{0} is continuous (as compositionof continuous maps). So it remains to show that F is ε− δ-continuousat x = 0. We note that

||F (x)− F (0)|| = ||F (x)|| = ||x|| = ||x− 0||

and hence for given ε > 0 we can choose δ = ε to obtain

||x− 0|| < δ =⇒ ||F (x)− F (0)|| < ε.

This shows that F is ε− δ-continuous at x = 0 and hence F is contin-uous. Moreover, F is a homeomorphism. This can be seen by arguingthat F is bijective and using the fact that the domain (resp. range) ofF is compact (resp. Hausdorff). Alternatively, it can be shown thatthe ‘radial extension’ of the inverse f−1 is the inverse of F .

Now we define H : B1 → B2 to be the composition h2 ◦ F ◦ h−11 . We

note that the restriction of H to ∂B1 is equal to

h2 ◦ f ◦ h−11 = h2 ◦ h−1

2 ◦ h ◦ h1 ◦ h−11 = h

as desired (here hi is really the restriction of hi to S1, but we’ve sup-pressed that in our notation), thus proving the claim.

# 2.4.2 To show that the Klein bottle K can be obtained by attachingtwo Mobius bands along their boundary, it suffices to find a Mobiusinside of K and to show that its complement is again a Mobius band.Consider our standard gluing scheme for the Klein bottle:

K = • a // •b

��• a //

b

OO

•

23

Inside of it we find the following Mobius band (given by the middlerectangle):

K = • a // •b1

��• c //

b3

OO

•b2

��• d //

b2

OO

•b3

��• a //

b1

OO

•

.

Removing the middle rectangle (i.e., removing a Mobius band from theKlein bottle), we are left with the upper and lower rectangle; gluingthose along their edge labeled ‘a’ (by putting the bottom rectangle ontop of the top rectangle), we obtain the following picture:

• d // •b3

��• a //

b1

OO

•b1

��• c //

b3

OO

•

This is a Mobius band; and so we see that removing the interior of aMobius band from K we obtain again a Mobius band. In other words,the Klein bottle K consists of two Mobius bands glued along theirboundary circle.

# 2.6.1 We need to show P#P ≈ K. Let P1, P2 be two copies ofthe projective plane and suppose that B1 ⊂ P1, B2 ⊂ P2 are disks. Toform the connected sum P1#P2, we glue P1 − intB1 and P2 − intB2

along their boundary circle. By Lemma 2.4.4 the spaces P1 − intB1

and P2 − intB2 are both Mobius bands. Hence the surface P1#P2

obtained by gluing these Mobius bands along their boundary circle ishomeomorphic to the Klein bottle by the previous homework problem.

# 2.6.2 By the previous homework problem, K ≈ P#P , and henceK#P ≈ P#P#P and K#K ≈ P#P#P#P .

24

The surface T#T that problem 2.6.2 asks to identify is already onthe list. I suspect that this is a typo, and what the author meant to dois to ask about K#T . By Lemma 2.6.5, P#T ≈ P#P#P and hence

K#T ≈ P#P#T ≈ P#P#P#P.

The surface Q shown in Figure 2.6.11 is the connected sum of four toriand a Klein bottle (note that if you remove the ‘obvious’ three tori andthe Klein bottle, you still have a torus!). Hence

Q ≈ K#T#T#T#T ≈ P#P#T#T#T#T.

Using Lemma 2.6.5 four times, we can replace each copy of T in theabove connected sum by two copies of P and hence we see that Qis homeomorphic to the connected sum of 10 copies of the projectiveplane.

# 1. We’ve proved before χ(S2) = 2, χ(T ) = 0 and χ(P ) = 1.We’ve also proved the following formula for the Euler characteristic ofconnected sums of surfaces:

χ(Q1#Q2) = χ(Q1) + χ(Q2)− 2

We claim: χ(T# . . .#T︸ ︷︷ ︸n

) = 2 − 2n. Proof by induction. This is true

for n = 1. Now assume that it is true for n ≤ k and lets considerχ(T# . . .#T︸ ︷︷ ︸

k+1

):

χ(T# . . .#T︸ ︷︷ ︸k+1

) = χ(T# . . .#T︸ ︷︷ ︸k

#T )

=χ(T# . . .#T︸ ︷︷ ︸k

) + χ(T )− 2 = (2− 2k) + 0− 2 = 2− 2(k + 1)

Similarly, we claim: χ(P# . . .#P︸ ︷︷ ︸n

) = 2− n. Proof by induction. This

is true for n = 1. Now assume that it is true for n ≤ k and lets considerχ(P# . . .#P︸ ︷︷ ︸

k+1

):

χ(P# . . .#P︸ ︷︷ ︸k+1

) = χ(P# . . .#P︸ ︷︷ ︸k

#P )

=χ(P# . . .#P︸ ︷︷ ︸k

) + χ(P )− 2 = (2− k) + 1− 2 = 2− (k + 1)

25

#2. We note that S2, T are orientable, while P is not. This impliesthat the connected sum T# . . .#T is orientable while the connectedsum P# . . .#P is not. In particular, none of the surfaces in the firstpart of the list (containing S2 and connected sums of T ’s) can be home-omorphic to any of the surfaces in the second part of the list (connectedsums of projective planes); argument: if a surfaces Q1, Q2 are homeo-morphic, one of them contains a Mobius band if and only if the otherdoes; in other words, Q1 is non-orientable if and only if Q2 is non-orientable.

Since the Euler characteristics of all the surfaces in the first part ofthe list are different from each other (by problem # 1), none of thesespaces can be homeomorphic to each other (note that P is a pattern ofpolygons on a surface Q, and Q′ is a surface homeomorphic to Q, thenthe images of vertices/edges/faces of P under the homeomorphism leadto a pattern P ′ on Q with obviously the same number of vertices, edgesand faces. In particular, the Euler characteristics Q and Q′ agree.

Similarly, the Euler characteristics of all the surfaces in the secondpart of the list are different from each other, none of these spaces canbe homeomorphic to each other.

#3. We want to show that if Q is a surface on the list provided by theclassification theorem, then also Q#P , Q#T and Q#K is homeomor-phic to a surface on the list.

First let us consider the case Q = S2; then

Q#P ≈ P Q#T ≈ T Q#K ≈ K ≈ P#P

shows that the desired statement holds in this case (the homeomor-phism K ≈ P#P comes from Lemma 2.6.4).

Now assume Q is a connected sum of k copies of the torus T . Thenapplying repeatedly the homeomorphism T#P ≈ P#P#P we obtain

Q#P = T# . . .#T︸ ︷︷ ︸k

#P ≈ T# . . .#T︸ ︷︷ ︸k−1

#P#P#P ≈ · · · ≈ T# . . .#T︸ ︷︷ ︸2k+1

Similarly, using K ∼= P#P we obtain

Q#K = T# . . .#T︸ ︷︷ ︸k

#K ≈ T# . . .#T︸ ︷︷ ︸k

#P#P

≈ T# . . .#T︸ ︷︷ ︸k−1

#P#P#P#P ≈ · · · ≈ T# . . .#T︸ ︷︷ ︸2k+2

It is obvious that Q#T = T# . . .#T︸ ︷︷ ︸k+1

is on the list.

26

Finally, let us discuss the case where Q is the connected sum of kcopies of P . Then obviously Q#P is the connected sum of k+ 1 copiesof P and hence on the list. Similarly, Q#K ≈ Q#P#P is a connectedsum of k + 2 copies of P and in particular on the list. Finally,

Q#T = P# . . .#P︸ ︷︷ ︸k

#T ≈ P# . . .#P︸ ︷︷ ︸k−1

#P#T

≈ P# . . .#P︸ ︷︷ ︸k−1

#P#P#P = P# . . .#P︸ ︷︷ ︸k+2

#5.2.1 Strictly speaking, this statement is wrong; a counterexampleis provided by the empty set, which is a perfectly good example of acompact smooth surface M of R3 (although not a particularly inter-esting one), which also can be covered by the single coordinate patchx : U →M (where U is the empty set).

We will prove that a non-empty compact smooth surface in R3 cannotbe the image of a single coordinate patch by contradiction. So letus assume that the smooth surface M ⊂ R3 is the image of a singlecoordinate patch x : U → M , whose domain U is some open subset ofR2. By Proposition 5.2.5(i), the map x is a homeomorphism betweenU and x(U) = M . In particular, the compactness of M implies thatU is compact, and hence by Heine-Borel U is a closed and boundedsubset of R2. The connectivity of R2 implies that the closed and opensubset U must be all of R2 (which it can’t be since it is bounded) orthe empty set (which is can’t be since then also M would be the emptyset). This is the desired contradiction.# 5.2.2 A coordinate patch x : U → R3 has to satisfy the followingthree conditions:

(1) It is injective;(2) It is smooth;(3) Its Jacobi matrix Dx has rank 2

The map x : R2 → R3, ( st ) 7→(

st

e3s−t

)is a coordinate patch, since it

is injective (we can recover both s and t from the image of ( st ) underx) and smooth (since its coordinate functions are smooth). Finally, itsJacobi matrix has the form

Dx =

1 00 1∗ ∗

,

where ∗ denotes some undetermined entry of the matrix (we couldcalculate it, but its value is irrelevant for determining the rank of Dx).

27

This implies that the two column vectors are linearly independent andhence Dx has rank 2 (independent of s, t).

The map

y : (R− {0})× (R− {0}) −→ R3 ( st ) 7→(s2

t21

)is not a coordinate patch, since it fails to be injective (replacing sand/or t by its negative doesn’t change its image under y).

Also

z : R2 −→ R3 given by ( st ) 7→(est

t4

)fails to be injective (since y(( s0 )) is independent of s) and hence it isno coordinate patch.

# 5.2.7 We want to show that an open subset N ⊂ M of a smoothsurface M ⊂ R3 is again a smooth surface. To show this, we need toconstruct for every point p ∈ N a coordinate patch x : U → N withp ∈ x(U). Our assumption that M is a smooth surface implies thatthere is a coordinate patch y : V → M with p ∈ y(V ). We note thaty might not be a coordinate patch for N , since y(V ) might not becontained in N . However, it is easy to fix this: y(V ) is an open subsetof M (by Prop. 5.2.5(i)), and hence so is y(V )∩N . By continuity of y,the preimage U := y−1(y(V ) ∩N) is an open subset of V . Restrictingthe coordinate patch y : V → M to U ⊂ V gives a map x : U → N ⊂M , which is again a coordinate patch (injectivity, smoothness and therank condition remain true after restricting the domain to some opensubset).

# 5.2.11 We want to prove Lemma 5.2.7, i.e., we want to show that ifx : U →M and y : V →M are coordinate patches then f ◦x : Axy → Ris smooth if and only if f ◦ y : Ayx → R is smooth, where

Axy = x−1(x(U) ∩ y(V )) Ayx = y−1(x(U) ∩ y(V )).

We recall that by Prop. 5.2.5(iii) the change of coordinate function

φx,y = y−1 ◦ x : Axy −→ Ayx

is a diffeomorphism; in particular, it is smooth. We note that

f ◦ x|Axy = f ◦ y ◦ y−1 ◦ x|Ax,y = f ◦ y ◦ φx,y |Axy

This implies that if f◦y|Ayx is smooth, then the composition f◦y◦φx,y =f ◦ x|Axy is smooth. Reversing the roles of x and y proves the oppositeimplication.

# 5.3.2 We want to show that the ellipsoid given by the equationx2

a2 + y2

b2+ z2

c2= 1 is a smooth surface. We note that the ellipsoid

28

is the preimage F−1(1) ⊂ R3 of the function F : R3 → R given by

F (x, y, z) = x2

a2 + y2

b2+ z2

c2. By Proposition 5.3.1 it suffices to show that

1 ∈ R is a regular value of F , i.e., that for all(xyz

)∈ F−1(1) the matrix

DF (p) =(

2xa2

2yb2

2zc2

)has rank one. We note that this matrix has rank zero if and only if all

its entries are zero which happens only for(xyz

)(which is not a point

of F−1(1)); in particular, 1 is a regular value of F and hence F−1(1) isa smooth surface.

# 5.3.3 To explicitly parametrize the hyperboloid x2

4+ y2

4− z2

9= 1,

we notice that it is invariant under rotations around the z-axis (since

the value of x2

4+ y2

4− z2

9does only depend on the norm of the vector

( xy )). This shows that the hyperboloid is obtained by rotating the

curve x2

4− z2

9= 1, x > 0 in the x-z-plane around the z-axis. There are

many ways to parametrize the hyperbola x2

4− z2

9= 1, x > 0; the most

common one is to use the hyperbolic sines and cosines and to set

c : R→ R2 c(t) =

(r(t)z(t)

)=

(2 cosh t3 sinh t

)Then as in formula (5.3.1) in the book, the surface of rotation can thenbe parametrized by

x : R2 −→ R3 given by x(t, θ) =

r(t) cos θr(t) sin θz(t)

=

2 cosh t cos θ2 cosh t sin θ

3 sinh t

# 5.4.3 Let M be the surface given by the graph of a smooth func-tion f : U → R. To compute the unit normal vector at a point p =( s

tf(s,t)

)∈M , we use the standard Monge coordinate patch

h : U −→M ⊂ R3 given by h(s, t) =

st

f(s, t)

Then

∂h

∂s=

01∂f∂s

and∂h

∂t=

10∂f∂t

29

and hence

∂h

∂s× ∂h

∂t=

∣∣∣∣∣∣e1 1 0e2 0 1

e3∂f∂s

∂f∂t

∣∣∣∣∣∣ =

−∂f∂s

−∂f∂t

1

It follows that

||∂h∂s× ∂h

∂t|| =

√(∂f

∂s

)2

+

(∂f

∂t

)2

+ 1

and hence the unit normal vector is given by

n =1√(

∂f∂s

)2+(∂f∂t

)2+ 1

−∂f∂s

−∂f∂t

1

.

#5.4.4 Let M be a surface of revolution and let h : J ×R→M ⊂ R3

be the standard parametrization of M given by (cf. equation 5.3.1 inthe book)

h(s, t) =

r(t) cos θr(t) sin θz(t)

Then

∂h

∂t=

r cos θr sin θz

and∂h

∂θ=

−r sin θr cos θ

0

and hence

∂h

∂t×∂h∂θ

=

∣∣∣∣∣∣e1 r cos θ −r sin θe2 r sin θ r cos θe3 z 0

∣∣∣∣∣∣ =

−zr cos θzr sin θ

rr cos2 θ + rr sin2 θ

=

−zr cos θzr sin θrr

It follows that

||∂h∂t× ∂h

∂θ||2 = z2r2 + r2r2

and hence the unit normal vector is given by

n =1√

z2 + r2

−z cos θz sin θr

.

30

# 5.4.6 Let M = F−1(a) be the level surface for some regular valuea ∈ R of a smooth function F : V → R. We want to show that thefunction n : M → R3 defined by

n =(DF )t

||(DF )t||is a unit normal vector field defined on all of M . We note that for everypoint p ∈M the derivative DF (p) is non-zero since a is assumed to bea regular value. In particular, ||DF (p)|| 6= 0 for all p ∈M and hence nis a unit vector field defined on all of M . It remains to be shown thatthe column vector (DF (p))t is perpendicular to the tangent space TpM .By definition, every tangent vector v ∈ TpM can be written in the formv = c(0), where c : (−ε, ε) → M is a smooth curve with c(0) = p forsome ε > 0. We note that then F (c(t)) = a for all t ∈ (−ε, ε) and hence

the derivative d(F◦c)dt

(t) = 0 for every t ∈ (−ε, ε). On the other hand,we can calculate the derivative of F ◦ c at t = 0 using the chain rule:

d(F ◦ c)dt

(0) = (DF )(c(0)) · c(0) = (DF )(p) · v = 〈(DF )t(p), v〉.

Here (DF )(p) · v is matrix product of the 1 × 3-matrix DF (p) (i.e.,row vector) and the 3× 1-matrix v (i.e., column vector), which can bealternatively be written as the scalar product 〈(DF )t(p), v〉 of the twocolumn vectors (DF )t(p), v. This shows that (DF )t(p) is perpendicularto any tangent vector v ∈ TpM .

# 5.4.7 Let M ⊂ R3 be a smooth surface, and let us assume thatfor a coordinate patch x : U →M (with U connected) the unit normalvector is given by

n(p) = kx(p) + w for all p ∈ U.Claim. If k = 0, then x(U) is contained in some plane perpendicularto w.

To prove the claim, it suffices to show that for any p, p′ ∈ U , thedifference x(p) − x(p′) is perpendicular to w or equivalently, that theinner product 〈x(p)− x(p′), w〉 vanishes. This in turn is equivalent tothe function

f : U −→ R given by p 7→ 〈x(p), w〉being a constant function on U . Following the hint in the book, we willprove this by showing that both partial derivatives of f are zero. Thisis based on the following lemma that we will prove below.

31

Lemma 1. Let f : U → R be a smooth function whose domain is openand connected. If all partial derivatives of f are zero, then f is aconstant function.

To apply this lemma to the function f above, we now calculate thepartials of f :

∂f

∂s=

∂

∂s〈x,w〉 = 〈∂x

∂s, w〉 = 0,

since ∂x∂s

is a tangent vector, while w = n is a normal vector. The sameargument shows that the other partial derivative of f vanishes as well,which implies that f is constant.

Claim. If k 6= 0, then x(U) is contained in a sphere of radius 1|k| with

center − 1|k|w.

We note that n is a unit vector and hence

||x(p)− (− 1

|k|w)|| = ||x(p) +

1

|k|w|| = || n

|k||| = 1

|k|||n|| = 1

|k|,

which proves the claim (in particular, we don’t need to calculate deriva-tives of this function as suggested in the hint to show that this functionis constant).

Proof of lemma. Let p0 = (s0, t0) and p1 = (s1, t1) be two points in Uand let us assume that the line segment connecting (s0, t0) and (s1, t0)as well as the line segment connecting (s1, t0) and (s1, t1) are containedin U . Then the Fundamental Theorem of Calculus implies

f(s, t) = f(s0, t0) +

∫ s1

s0

∂f

∂s(s, t0) ds+

∫ t1

t0

∂f

∂t(s1, t) dt = f(s0, t0).

This shows that f is locally constant in the sense that for every pointp ∈ U , there is an open neighborhood V such that f restricted to V isconstant.

We prove that f is constant on U by contradiction. Assume thatthere are points p0, p1 ∈ U with f(p0) 6= f(p1). Define subsets A,B ⊂U as follows:

A := {p ∈ U | f(p) = f(p0)} B := {p ∈ U | f(p) 6= f(p0)}.

By construction, A, B are non-empty, and their union is U . Both ofthese sets are open in U since f is locally constant. This is a contra-diction to our assumption that U is connected. �

32

# 5.5.1 Using problem # 5.4.3 and adopting the suggested notationf1 = ∂f

∂s, f2 = ∂f

∂twe have

g11 = 〈∂h∂s,∂h

∂s〉 = 〈

01f1

,

01f1

〉 = 1 + (f1)2

g12 = g21 = 〈∂h∂s,∂h

∂t〉 = 〈

01f1

,

10f2

〉 = f1f2

g22 = 〈∂h∂t,∂h

∂t〉 = 〈

10f2

,

10f2

〉 = 1 + (f2)2

Hence in matrix notation we have

(gij) =

(1 + (f1)

2 f1f2

f1f2 1 + (f2)2

).

# 5.5.5 Using problem # 5.4.4 we have

g11 = 〈∂h∂t,∂h

∂t〉 = 〈

r cos θr sin θz

,

r cos θr sin θz

〉= r2 cos2 θ + r2 sin2 θ + z2 = r2 + z2

g12 = g21 = 〈∂h∂t,∂h

∂θ〉 = 〈

r cos θr sin θz

,

−r sin θr cos θ

0

〉= −r cos θr sin θ + r sin θr cos θ = 0

g22 = 〈∂h∂θ,∂h

∂θ〉 = 〈

−r sin θr cos θ

0

,

−r sin θr cos θ

0

〉= r2 sin2 θ + r2 cos2 θ = r2

Hence in matrix notation we have

(gij) =

(r2 + z2 0

0 r2

).

33

# 5.7.3 For this and the following problem, we will use equation (5.7.4)from the book, i.e.: (

Γ111

Γ211

)= g−1

(12∂g11∂u1

∂g12∂u1− 1

2∂g11∂u2

)(

Γ112

Γ212

)= g−1

(12∂g11∂u2

12∂g22∂u1

)(

Γ122

Γ222

)= g−1

(∂g12∂u2− 1

2∂g22∂u1

12∂g22∂u2

)We recall from problem # 5.5.5 that

g =

(r2 + z2 0

0 r2

)and hence g−1 =

((r2 + z2)−1 0

0 r−2

)and(

12∂g11∂u1

∂g12∂u1− 1

2∂g11∂u2

)=

(rr + zz

0

) (12∂g11∂u2

12∂g22∂u1

)=

(0rr

) (∂g12∂u2− 1

2∂g22∂u1

12∂g22∂u2

)=

(−rr

0

)Hence

Γ111 =

rr + zz

r2 + z2Γ2

12 =r

rΓ1

22 = − rr

r2 + z2

and all other Christoffel symbols vanish.

Now let us assume that the curve c(t) =

(r(t)0z(t)

)has unit speed

parametrization. Then

1 = ||c||2 = r2 + z2,

and hence

0 =∂

∂t

(r2 + z2

)= 2rr + 2zz.

This simplifies the above results to

Γ212 =

r

rΓ1

22 = −rr

and all other Christoffel symbols vanish.

# 5.7.4 For

g =

(1 00 G

)and hence g−1 =

(1 00 G−1

)

34

and(12∂g11∂u1

∂g12∂u1− 1

2∂g11∂u2

)=

(00

) (12∂g11∂u2

12∂g22∂u1

)=

(0

12∂G∂u1

) (∂g12∂u2− 1

2∂g22∂u1

12∂g22∂u2

)=

(−1

2∂G∂u1

12∂G∂u2

)Hence

Γ212 = Γ2

21 =1

2G

∂G

∂u1

Γ122 = −1

2

∂G

∂u1

Γ222 =

1

2

∂G

∂u2

and all other Christoffel symbols vanish.

# 5.8.2 To calculate the length of the curve

c : [1, 1] −→ R3 c(t) =

et

et

e2t

,

we first determine the tangent vector to be

c(t) =

et

et

2e2t

and hence ||c(t)|| =√

2e2t + 4e4t

Then

length(c) =

∫ 1

−1

||c(t)|| dt =

∫ 1

−1

√2e2t + 4e4t dt =

∫ 1

−1

et√

2 + 4e2t dt

Substituting x = 2et, this is equal to

1

2

∫ 2e

2e

√2 + x2 dx

Now using the integration formula (e.g., back of Thomas’ Calculus),∫ √a2 + x2 dx =

x

2

√a2 + x2 +

a2

2ln(x+

√a2 + x2) + C

we obtain

length(c) =1

2

[x2

√2 + x2 + ln(x+

√2 + x2)

]2e2e

=e

2

√2 + 4e2 +

1

2ln(2e+

√2 + 4e2)− 1

2e

√2 +

4

e2− 1

2ln(

2

e+

√2 +

4

e2)

# 5.8.3 We parametrize the torus of large radius R and small radius a(to avoid a conflict of notation) as a surface of revolution obtained by

35

rotating the circle of radius a around the point ( R0 ) around the z-axis.This circle can be parametrized by

c : [0, 2π] −→ R3 c(t) =

r(t)0z(t)

=

R + a cos t0

a sin t

and hence the torus by (see equation 5.3.1 in the book)

x : [0, 2π]× [0, 2π] −→ T x(t, θ) =

r(t) cos θr(t) sin θz(t)

According to problem # 5.5.5, the associated matrix coefficients areequal to

g =

(r2 + z2 0

0 r2

)=

((−a sin t)2 + (a cos t)2 0

0 (R + a cos t)2

)=

(a2 00 (R + a cos t)2

)and hence

√det g = a(R + a cos t). It follows that

area(T ) =

∫ 2π

0

∫ 2π

0

√det g dt dθ = 2πa

∫ 2π

0

(R + a cos t)dt

= (2πa)(2πR) + 2πa2

∫ 2π

0

cos tdt = 4π2aR

# 5.8.5 According to problem # 5.5.1, for a monge patch f : U → R,the associated matrix coefficients are equal to

g =

(1 + f 2

1 f1f2

f1f2 1 + f 22

)and hence

det g = (1 + f 21 )(1 + f 2

2 )− f 21 f

22 = 1 + f 2

1 + f 22 .

If R ⊂ U , then

area(graph(f|R)) =

∫R

√det g ds dt =

∫R

√1 + f 2

1 + f 22 ds dt.