MA-643 (Algebraic Topology) Assignment · 2015-03-03 · Assigned Problem Solution 3. 132123005 (AYAN SENGUPTA) 3.1Assigned Problem Problem-1 Problem-2 Problem-3 (7, 4.3) (46, 4.8)

MA-643 (Algebraic Topology) Assignment

Department of Mathematics

Indian Institute of Technology Guwahati—————————————–

Contents

1 07012321 (SUBHASH ATAL) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.1 Assigned Problem 5

1.2 Solution 5

2 11012116 (GIRIDHAR V. KULKARNI) . . . . . . . . . . . . . . . . . . . . . 6


2.2 Solution 6

3 132123005 (AYAN SENGUPTA) . . . . . . . . . . . . . . . . . . . . . . . . . . . 7


3.2 Solution 7

4 132123009 (BILTU DAN) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11


4.2 Solution 11

5 132123011 (DIKSHYA RATHA) . . . . . . . . . . . . . . . . . . . . . . . . . . . 13


5.2 Solution 13

3

6 132123017 (MOHIT TRIPATHI) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15


6.2 Solution 15

7 132123019 (PALASH SARKAR) . . . . . . . . . . . . . . . . . . . . . . . . . . . 16


7.2 Solution 16

8 132123025 (RAKESH JANA) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17


8.2 Solution 17

9 132123033 (SOMNATH GHOSH) . . . . . . . . . . . . . . . . . . . . . . . 20


9.2 Solution 20

10 132123036 (SUBHA PAL) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22


10.2 Solution 22

11 132123040 (SUJEET KR. SINGH) . . . . . . . . . . . . . . . . . . . . . . . . . 23


11.2 Solution 23

12 132123041 (SUMIT DAS) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25


12.2 Solution 25

NOTE

All problems are taken from the book A Basic Course in Algebraic Topology by W.M.Massey.

(M,T.S) ; M,T,S ∈N where M represents page number, T represents section numberand S represents problem number.

Assigned problems are as follows :

Roll No Name Problem-1 Problem-2 Problem-307012321 SUBHASH ATAL (7, 4.1) (46, 4.6) (53, 7.3)11012116 GIRIDHAR V. KULKARNI (7, 4.2) (46, 4.7) (53, 7.4)132123005 AYAN SENGUPTA (7, 4.3) (46, 4.8) (53, 7.5)132123009 BILTU DAN (13, 5.1) (46, 4.9) (53, 7.6)132123011 DIKSHYA RATHA (42, 3.1) (46 4.10) (57, 8.1)132123017 MOHIT TRIPATHI (42, 3.2) (46, 4.11) (58, 8.2)132123019 PALASH SARKAR (42, 3.3) (49, 5.1) (66, 3.1)132123025 RAKESH JANA (43, 4.1) (50, 5.2) (69, 3.2)132123033 SOMNATH GHOSH (44, 4.2) (50, 5.3) (70, 3.3)132123036 SUBHA PAL (45, 4.3) (50, 5.4) (70, 3.4)132123040 SUJEET KUMAR SINGH (45, 4.4) (53, 7.1) (71, 3.5)132123041 SUMIT DAS (45, 4.5) (53, 7.2) (74, 4.1)

√means submitted.

× means not submitted.

Assigned ProblemSolution

1. 07012321 (SUBHASH ATAL)

1.1 Assigned ProblemProblem-1 Problem-2 Problem-3

(7, 4.1) (46, 4.6) (53, 7.3)× × ×

1.2 SolutionNot submitted


2. 11012116 (GIRIDHAR V. KULKARNI)


(7, 4.2) (46, 4.7) (53, 7.4)× × ×

2.2 SolutionNot submitted


3. 132123005 (AYAN SENGUPTA)


(7, 4.3) (46, 4.8) (53, 7.5)√ √ √

3.2 SolutionQuestion 4.3 (Page no. 7):If S1 and S2 are projective planes, then S1#S2 is a "Klein Bottle" i.e. homeomorphic tothe surface obtained by identifying the opposite sides of a square.

Solution: We can prove this by using cut and paste technique.We can identify Klein Bottle as follows :

We identify projective planes as follows :

As, the connected sum does not depend on the choice of hole, we cut the followingcircular hole from each of the projective planes

8 Chapter 3. 132123005 (AYAN SENGUPTA)

After deleting the interior of the hole what we get is a mobius strip with boundary.

Now we will see that Klein Bottle is an union of two mobius strips.

Now, if we cut out the 2nd portion and paste the 3rd portion and 1st portion along theidentification "a", then we get two mobius strips with boundaries.

Hence, union of two mobius strips gives us Klein Bottle. So, connected sum twoprojective planes is Klein Bottle. �

3.2 Solution 9

Question 4.8 (Page no. 46)Let T be a torus and let X be the complement of a point in T. Find a subset of X whichis homeomorphic to a figure "8" curve (i.e. the union of two circles with one pointcommon) and which is deformation retract of X .

Solution: First consider the torus -

It is clear from the above pictures that the boundary of the torus is homeomorphic to thefigure 8. Now we will show that this is a deformation retract of a punctured torus.Consider the square S = [−1,1]× [−1,1] with the above identifications and take T =S− (0,0) be the punctured torus. Consider F : T × [0,1]→ T

F((x,y), t) = ((1− t)x,(1− t)y)+ (tx,ty)max{|x|,|y|} .

It is clear that F is continuous in T × [−1,1]. Moreover,F((x,y),0) = (x,y)F((x,y),1) = (x,y)

max{|x|,|y|} which is on the boundary of T .For (x,y) ∈ δT , max{|x|, |y|}= 1. Hence,F((x,y), t) = ((1− t)x,(1− t)y)+(tx, ty) = (x,y). So, δT is a deformation retract of Twhich is homeomorphic to the figure 8. �

10 Chapter 3. 132123005 (AYAN SENGUPTA)

Page 53: Problem 7.5Assume that G is a topological space, µ : G×G→ G is a continuous map, and e ∈ Gis such that the following conditions hold: For any x ∈ G, µ(x,e) = µ(e,x) = x.[An important example: G is a topological group, e is the identity element, andµ(x,y) = xy for all x,y ∈ G]. Let i : G→ G×G and j : G→ G×G be defined as:i(x) = (x,e) and j(x) = (e,x) for any x ∈ G. Prove that, for any element β ,γ ∈ π(G,e),µ∗[(i∗β ).( j∗γ)] = β .γ . Deduce as a corollary that π(G,e) is an abelian group.

Solution : Let β = [α1] i.e. class of loops in G at point e which are homotopic tothe loop α1, whereα1 : [0,1]→ G such that α1(0) = e = α1(1).Similarly, γ = [α2], where α2 : [0,1]→ G such that α2(0) = e = α2(1).Now, i∗ : π(G,e)→ π(G×G,(e,e)) such that i∗([α]) = [i◦α]. Similarly j∗ : π(G,e)→π(G×G,(e,e)) such that j∗([α]) = [ j ◦α].Hence, i∗([α1]) = [i◦α1] = [(α1(t),e)] = i∗β and j∗([α2]) = [ j◦α2] = [(e,α2(t))] = j∗γ .Hence, [(i∗β ).( j∗γ)] = [α3], where

α3 = (α1(t),e).(e,α2(t)) =(α1(2t),e) 0≤ t ≤ 1/2

(e,α2(2t−1)) 1/2≤ t ≤ 1

Now, µ∗([α3]) = [µ ◦α3] = [µ((α1(t),e).(e,α2(t)))] = [µ((α1(t),e)).µ((e,α2(t)))] =[α1(t).α2(t)] = β .γFrom the previous problem we have got that the elements i∗β and i∗γ commutes. i.e.(i∗β ).( j∗γ) = ( j∗γ).(i∗β ).Hence,β .γ = µ∗[(i∗β ).( j∗γ)] = µ∗[( j∗γ).(i∗β )] = γ.β . This is true for any β ,γ ∈ π(G,e).Hence, we have proved that π(G,e) is an abelian group. �


4. 132123009 (BILTU DAN)


(13, 5.1) (46, 4.9) (53, 7.6)√ √ √

4.2 SolutionPage No.: 13, Question No.: 5.1Question: Let P be a polygon with an even number of sides. Suppose that the sides areidentified in pairs in accordance with any symbol whatsoever. Prove that the quotientspace is a compact surface.

Answer: Let the number of sides of the polygon P be 2n and let X be the quotientspace obtained by identifying the sides of P in pairs.We prove that X is a compact surface.We first prove that every point of X has a neighborhood homeomorphic to the unit openball centered at origin in R2 . A point in the X either corresponds to one point p in theinterior of P, to two points q1,q2 on identified edges (but not vertices) of P, or to severalvertices r1,r2, . . . ,rk (k ≤ 2n) of P.In the first case, we can take our open set to be a ball around p that does not reach theboundary of P.In the second case, a neighborhood can be taken to be a half-disk around q1 and ahalf-disk around q2 , joined on their common diameter to make a whole disk.In the third case, if k = 1 then a neighborhood of r1 is a pie wedge with edges identified,i.e. a cone, which is homeomorphic to the disk.If k ≥ 2, we consider the edge parts that run into or out of the various vertices ri (forexample, front of edge a, back of edge a, front of edge b, etc.). Each part occurs exactlytwice in P, and hence exactly twice in the list. We call these edge parts e1,e2, . . . ,ek ,and label them (and relabel the vertices) so that r1 is connected to edge parts e1 and e2

12 Chapter 4. 132123009 (BILTU DAN)

,r2 is connected to e2 and e3 , . . .,rk−1 is connected to ek−1 and ek , and rk is connectedto ek and e1. A neighborhood of our point in X now corresponds to a bunch of piewedges, sandwiched between edges e1 and e2, e2 and e3 , etc. But as each wedge ishomeomorphic to a pie wedge of angle 2π/k, so their union is homeomorphic to anentire pie ( i.e., to an open ball in R2 ) and hence to the open unit ball in R2 .Now we notice that given two distinct points in X , the neighbourhoods we have justconstructed can be taken small enough so that they do not overlap. Therefore the spaceX is Hausdorff.Again since we are taking the quotient of a compact set (a closed polygon) by someidentifications. So X is compact.Thus X is a compact surface.

Page No.: 46, Question No.: 4.9Question: Let S be a compact surface and let X be the complement of a point in S. Finda subset A of X such that(a) A is homeomorphic to the union of a finite number of circles and(b) A is a deformation retract of X .

Answer: Let us assume that the space S is obtained as quotient space by identifyingthe sides in pairs of a polygon P and suppose that the origin lies in the interior of P. Letthe origin be the point that is removed.Note that for any point in S we can make thisassumption. Now we take A to be the subset of X corresponding to the boundary of(with the usual identifications) of P. The retraction map is just moving radially outwardsuntil we hit the boundary.It is easy to see that {1} is a deformation retract of (0,1].Withthis observation we can conclude that the boundary of P is a deformation retract of thepunctured polygon P\{(0,0)} and hence A is a deformation retract of X .Clearly A is homeomorphic to the union of a finite number of circles.


5. 132123011 (DIKSHYA RATHA)


(42, 3.1) (46 4.10) (57, 8.1)√ √ √

5.2 SolutionQ1. Under what conditions will two path classes γ and γ ′ from x to y give rise to the

same isomorphism of π(X ,x) and π(X ,y)?Soln. Suppose that for γ and γ ′ connecting x to y we have the same isomorphism.

Then for every φ ∈ π(X ,x), we have (γ)1φγ =(γ ′)1φγ ′ which implies γ(γ)1φγγ ′1 =φ Consequently γγ1 = (γ(γ)1)1andγ ′(γ)1 ∈ π(X ,x). Furthermore the equationabove implies that γ(γ)1 belongs to the center. And the converse is evident.We have got a criterion: γ and γ ′ rise the same isomorphism iff γ(γ)1 belongs tothe center of π(X ,x).

Q2. x ,y be points of a simply connected space X.Prove that there exists a unique pathclass in X with initial point x and terminal point y.

Soln. Let α and β be two paths from x to y. Then α ∗β−1 is defined and is a loop basedon X at x.Since X is simply connected,this loop is path connected to the constant loop at x.Then,[α ∗β−1]∗ [β ] = [ex]∗ [β ], from which it follows that [α] = [β ].

Q3. Prove that if A is a deformation retract of X,then the inclusion ı : A 7−→ X is ahomotopy equivalence.

Soln. A is a deformation retract of X iff exists a retraction r : X → A and a homotopyf : X× I→ X such thatf (x,0) = xf (x,1) = r(x), x ∈ X ,and f (a, t) = a, a ∈ A, t ∈ I.

14 Chapter 5. 132123011 (DIKSHYA RATHA)

Since r(a) = a ∀a ∈ A (retraction), so r ◦ ı = 1A, while f is an homotopy between1X and ı◦ r, so r and ı are homotopy equivalences.


6. 132123017 (MOHIT TRIPATHI)


(42, 3.2) (46, 4.11) (58, 8.2)√× ×

6.2 SolutionPage no. 42 , Question:3.2 :- Let X be an arcwise-connected space.Under what condi-tions is the following statement true:For any two points x,y∈X,all path classes from x toy give rise to the same isomorphism of π(X ,x) onto π(X ,y).

Answer: The path from x to y does not matter if π is Abelian. The path does matter ifπ is non-Abelian. To see this, note that the difference between the map induced by twopaths γ,γ

′from x to y is conjugation by γ−1γ

′. Specifically, if

f ,g : π(X ,x)−→ π(X ,y)

are the two maps, then for each α ∈ π(X ,x), f (α) = γ−1γ′g(α)γ

′−1γ. If π is abelian,this is the same as g(α). If π is not abelian, we can take y = x to be the identity and γ

′

to be the element of π that is not in the center of the group.


7. 132123019 (PALASH SARKAR)


(42, 3.3) (49, 5.1) (66, 3.1)√× ×

7.2 SolutionQuestion:Let f ,g : I−> X be two paths with initial point x0 and end point x1 . provethat f ∼ g iff f g′ ∼ e at x0.

ANSWER:We know that if f ,g,h, t : I−> X are four paths and f ∼ g and h ∼ t thenf h∼ gt. with same starting and end points then Given that f ∼ g. Now is very trivialthat g′ ∼ g′ and hence f g′ ∼ gg′ .Since gg′ ∼ e , f g′ ∼ e.Converseley,assume that f g′ ∼ e.Since g ∼ g we have f g′g ∼ eg since f g′g ∼ f ndeg ∼ g. we have f ∼ g. since f andg aretwo paths with same starting and end points,hence f g′ will be a loop based at x0.


8. 132123025 (RAKESH JANA)


(43, 4.1) (50, 5.2) (69, 3.2)√ √ √

8.2 SolutionProblem — (43,4.1). Let ϕ : X → Y be a continuous map and let γ be a class ofpaths in X from x0 to x1. Prove that the following diagram is commutative:

Π(X ,x0) Π(Y,ϕ(x0))

Π(X ,x1) Π(Y,ϕ(x1))

u

ϕ∗

ϕ∗

v

Here the isomorphism u is defined by u([ f ]) = [γ−1] ∗ [ f ] ∗ [γ] and v is definedsimilarly using ϕ∗([γ]) in place of [γ].

Answer: Let u is path in X from x0 to x1 and v = ϕ ◦ u. In order to show givendiagram is commutative, we have to show v◦ϕ∗ = ϕ∗ ◦ u. Thus we need to show thatthese two operation do the same thing to any [ f ] ∈Π(X ,x0).Let f be any path in X based at x0. Then we have

(v◦ϕ∗)([ f ]) = (ϕ ◦u)◦ϕ∗([ f ]) by def of v

= (ϕ ◦u)([ϕ ◦ f ]) by def of ϕ∗

= [(ϕ ◦u)−1]∗ [ϕ ◦ f ]∗ [ϕ ◦u] by def of v

18 Chapter 8. 132123025 (RAKESH JANA)

Again now,

(ϕ∗ ◦ u)([ f ]) = ϕ∗([u−1]∗ [ f ]∗ [u]) by def of u

= ϕ∗([u−1 ∗ f ∗u]) by def of ∗= [ϕ ◦ ([u−1 ∗ f ∗u])] by def of ϕ∗

= [(ϕ∗ ◦u−1)∗ (ϕ ◦ f )∗ (ϕ ◦u) k ◦ ( f ∗g) = (k ◦ f )∗ (k ◦g)

= [ϕ∗ ◦u−1]∗ [ϕ ◦ f ]∗ [ϕ ◦u] by def of ∗

Now (ϕ ◦u)−1(t) = (ϕ ◦u)(1− t) = ϕ(u(1− t)) = (ϕ ◦u−1)(t), for any t ∈ [0,1].Hence the result. �

Corollary: Suppose X = U ∪V , where U and V are open sets of X ; Suppose U ∩Vis nonempty and path connected. If U and V are simply connected, then X is simplyconnected.

Problem — (50,5.2). Prove that the unit-2 sphere S2 ,or more generally, unit-nsphere Sn n≥ 2 is simply connected.

Answer: Let p = (0, . . . ,1) ∈ Rn+1 and q = (0, . . . ,−1) ∈ Rn+1 be the "north pole"and the "south pole" of Sn,respectively.Claim: For n≥ 1, the punctured sphere Sn− p is homeomorphic to Rn.Define f : (Sn− p)→ Rn by the equation

f (x) = f (x1, . . . ,xn+1) =1

1− xn+1(x1, . . . ,xn).

The map f is called stereographic projection. If one takes the straight line in Rn+1

passing through the north pole p and the point x of Sn− p, then this line intersect then-plane Rn×0⊂ Rn+1 in the point f (x)×0. clearly if we define g : Rn→ (Sn− p)by the equation

g(y) = g(y1, . . . ,yn+1) = (t(y) · y1, . . . , t(y) · yn,1− t(y)),

where t(y) = 21+‖y‖2 , is a right and left inverse for f . Hence the claim.

Claim: Sn− p and Sn−q are homeomorphic and hence both are isomorphic to Rn.It is easy to check. Take the reflection map (x1, . . . ,xn+1)→ (x1, . . . ,xn,−xn+1).Now take U = Sn− p and V = Sn− q be two open set of Sn. Note that U and Vare path connected because they are homeomorphic image of Rn and have a pointcommon (1,0, . . . ,0) of Sn.Hence Sn is path connected for n≥ 1.The space U and V are simply connected, being homeomorphism to Rn. Theirintersection equals Sn− p−q, which is homeomorphic under stereographic projectionto Rn−0. The latter space is path connected, for every point Rn−0 can be joined toapoint of Sn−1 by a straight line path, and Sn−1 is path connected for n≥ 2. Hence theresult using the above corollary. �

Problem — (69,3.2). Let S is a finite set and S′ is an infinite set, and let F and F ′ arefree abelian group on S and S′, respectively. Show that F and F ′ are non-isomorphic.

8.2 Solution 19

Answer: The proof is by contradiction. If possible let F and F ′ are isomorphic. Thenany isomorphism between F and F ′ induces an isomorphism between the quotientgroups F/Fn and F ′/F ′n. But the quotient group F/Fn is a finite group of ordernk, where k is cardinality of S and the quotient group F ′/F ′n is an infinite group.Hence we can not get any isomorphism between a finite group and an infinite group, acontradiction. Hence the result.

�


9. 132123033 (SOMNATH GHOSH)


(44, 4.2) (50, 5.3) (70, 3.3)√ √ √

9.2 SolutionPage 44, Question No: 4.2Question: Show that a retract of a Hausdorff space must be a closed subset.Answer: Let X be a Hausdorff space and A⊂ X be a retract of x. Let r : X → A be theretraction mapping. To show that, A is closed set. For this we show that X \A is an opensubset of X .If A = X , nothing to prove, let X \A 6= φ . Take x ∈ X \A. Then x 6= r(x), as r(x) ∈ A.Therefore x and r(x) are two different elements of the Hausdorff space X . Then thereexists two open sets U and V in X such that x∈U and r(x)∈V . Then x∈ r−1(V ∩A), sox∈ (r−1(V ∩A))∩U . Since r is continuous r−1(V ∩A) is open in X, so (r−1(V ∩A))∩Uis open in X . We want to show that x is an interior point of X \A. Suppose there is a pointa ∈ A such that a ∈ (r−1(V ∩A))∩U . Then a ∈ r−1(V ∩A), this implies r(a) ∈ (V ∩A).Since a ∈ A implies r(a) = a ∈V . Thus a ∈V ∩U , which is a contradiction. Therefore,x is an interior point of X \A, that is X \A is open. Therefore A is a closed set.

Page 50, Question No: 5.3Question: Prove that R2 and Rn are not homeomorphic if n 6= 2.Answer: Suppose f : R2→ Rn is a homeomorphism.If n=1: Then f induces a homeomorphism between R2 \{(0,0)} and R\{ f ((0,0))}.But R2 \{0} is connected and R\{ f (0)} is not connected. Therefore there is no home-omorphism between them, hence no such f exists.If n≥ 3: Then f induces a homeomorphism between R2 \{(0,0)} and Rn \{ f ((0,0))}.

9.2 Solution 21

Since S1 is a deformation retract of R2 \ {(0,0)} and fundamental group of S1 is Z,therefore fundamental group of R2 \{(0,0)} is isomorphic to Z. But fundamental groupof Rn \{ f ((0,0))} is trivial, since it has a deformation retract to Sn−1 which has trivialfundamental group (for n≥ 3). But trivial group is not isomorphic to Z.Therefore there is no homeomorphism between R2 and Rn if n 6= 2.

Page 70, Question No: 3.3: Out of syllabus.


10. 132123036 (SUBHA PAL)


(45, 4.3) (50, 5.4) (70, 3.4)× × ×

10.2 SolutionNot Submitted


11. 132123040 (SUJEET KR. SINGH)


(45, 4.4) (53, 7.1) (71, 3.5)√ √ √

11.2 SolutionPage no. 45Question:4.4 Let A be a subspace of X, and Y be a non empty topological space. Provethat A × Y is a retract of X×Y iff A is a retract of X.

Answer - Suppose that A is a retract of X.Then, we need to prove that A×Y is aretract of X×Y.Since, A is a retract of X. ∃ a function

r : X −→ A

such that r(a)=a ∀a ∈ A and r is continous.Define

f : X×Y −→ A×Y,

f (x,y) = (r(x),y).

Since, r is continous f is continous and it satisfies all the condition of retract map fromA×Y to X×Y .

Now, suppose that A×Y be retract of X×Y.Therefore,there exists

h : X×Y −→ A×Y

such that h(a,y)=(a,y) ∀(a,y) ∈ A×Y .Let y0 ∈ Y , define

s : X −→ X×Y

24 Chapter 11. 132123040 (SUJEET KR. SINGH)

by s(x)=(x,y0) .Consider the function. t = πohos : X −→ A where π is the projection map from A×Yto A.Since s,h,π all are continous functions ,their composition will be also continous.Hence, t is continous .Also

t(a) = a. ∀a ∈ A

Therefore, t is a retraction map from X to A.Hence,A is retract of X.

Page no. 53Question:7.1 Describe the structure of the fundamental group of torus.Answer:We know if X is path connected and Y is path connected than

π(X×Y ) = π(X)×π(Y ).

Also, if X and Y are homemorphic than

π(X) = π(Y ).

We know that torus is homemorphic to S1×S1.Therefore,

π(T 2) = π(S1×S1).

= π(S1)×π(S1).

= Z×Z.

Since,π(S1) = Z.

Page no. 71Question:3.5 Give an example to show that in theorem 3.7 the subset S⊂F and thesubgroup F

′may be disjoint even in the case where the cardinals of S and S

′are equal.

Answer: Let S={x,y}.Than, the free abelian group of S call it, F=< x >×< y > .where, < t >={.......t−2, t−1, t0, t1, t2.....}w Z.Let F

′=< x2 >×< y5 > w 2Z×5Z.

Then, S′={x2,y5}.i.e F

′is the free abelian group of S

′.

NowS∩F

′= φ .

i.e there is no element S in F′even though the cardinality of S and S

′are equal.(NOTE:

here we are taking the natural mapping i from S to F i.e i(x)=x×y0 and i(y)=x0× y andin that sense we are taking intersection).


12. 132123041 (SUMIT DAS)


(45, 4.5) (53, 7.2) (74, 4.1)√ √ √

12.2 SolutionPage 45;Question no:4.5Question:Prove that the relation "is a retract of" is transitive i.e, if A is a retract of Band B is a retract of C, then A is a retract of C.

Answer: Here B is a retract of C, implies B⊂C.So, we have a retract r1 : C→ B with injection i1 : B→C such that r1 ◦ i1 = 1BAlso, A is a retract of B implies, A⊂ BSo, we have a retract r2 : B→ A with injection i2 : A→ B such that r2 ◦ i2 = 1A.We wantto show that we can retract C directly onto A.Note that, r2 ◦ r1 : C→ A and i1 ◦ i2 : A→C, we can see that

r2 ◦ r1 ◦ i1 ◦ i2 = r2 ◦ (r1 ◦ i1)◦ i2= r2 ◦ (1B)◦ i2= r2 ◦ i2= 1A

Thus r2 ◦ r1 is a retract from C to A.So, A is a retract of C.

Page 53;Question no:7.2Question:Prove that the subset S1× x0 is a retract of S1×S1, but that is not a deforma-tion retract of S1×S1 for any point x0 ∈ S1.

26 Chapter 12. 132123041 (SUMIT DAS)

Answer:Suppose x0 is any point of S1

Let us define the map r : S1×S1→ S1× x0 by r(x,y) = (x,x0) where x,y ∈ S1.This map is continuous and fixes pointwise the subspace S1× x0Let i : S1× x0→ S1×S1 be the inclusion map and

r ◦ i(x,x0) = r(i(x,x0))

= r(x,x0)

= (x,x0)

Hence, r is a retraction from S1×S1 to S1× x0.Deformation retractions are homotopy equivalences and induce isomorphisms of funda-mental groups.However S1× x0 is homeomorphic to S1 which has fundamental group z,whereas S1×S1 has fundamental group z×z.z is cyclic but z×z is not cyclic.So these twogroups are not isomorphic, So there can be no deformation retract of S1×S1 onto S1×x0.

Page 74;Question no:4.1Question:Let {Gi : i ∈ I} be a collection containing more than one group,each of whichhas more than one element.Prove that their free product is non-abelian,contains elementsof infinite order,and that its center consists of the identity element alone

. Answer:Let a ∈ G1 and b ∈ G2 be non-trivial elements.ab and ba are both in re-duced form,and are different words,So the free group is non-abelian.Also, (ab)n =abab....ab 6= 1, So ab has infinite order.Let W = g1g2....gn be an arbitrary non-emptyword,with g1 ∈Gi and gn ∈G j.If Gi 6=G j, then g−1 1W = g2....gn 6=Wg−1 1= g1....gng−1 1,so W is not in the center.If i = j, then pick a non-trivial element c ∈ Gk with k 6= i, andnote that cW 6=Wc.Thus W is not in the center of the group.

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MA-643 (Algebraic Topology) Assignment · 2015-03-03 · Assigned Problem Solution 3. 132123005 (AYAN SENGUPTA) 3.1Assigned Problem Problem-1 Problem-2 Problem-3 (7, 4.3) (46, 4.8)