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Combinatorial Topology (MAS430)

Homework 1

Duong Q. Dinh - 20121095

1 Problem 5, 4.2, Armstrongs Basic Topology

Solution:

To show X and Yare not homeomorphic, we will show that Xis not compact while

Y is not.

Let{U}A be an open cover ofX. Since(0, 0)X, there is some 0A such that(0, 0)U0. Then U0 =X E, where Eis an open subset ofE2. It is clear that onlya finite number of circles{C}B are not contained completely in U0 . For each B ,the part of the circle C not contained inU0 is homeomorphic to a closed interval in the

real line, and is thus compact. Then, from the open cover{U C}A ofC, we canextract a finite subcover ofC, say{UC}TA. It is easy to see that{U0, U}TA,

whereT = BT, is a finite cover ofX. Therefore X is compact.

Note that we can also prove the compactness ofXby showing its boundedness and

closedness in E2.

Let be the canonical projection to the quotient space Y = R/Z. Consider the

open subset N =

zZ

(z 1/2, z+ 1/2) of R. Since simply identifies the integerstogether, 1(N) = N. Since R/Z has the quotient topology, (N) is open. Let

Mz = (z+ 1/4, z+ 3/4)for z Z. Clearly then{(N)}{(Mz)}zZ is an open cover ofR/Z. However, we cannot extract any finite subcover out of this, since for anyz,z+ 1/2

is contained in only one open subset of form (z+ 1/4, z+ 3/4). Thus Y is not compact.


Solutions:

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We use the following lemma: If f : X Y is a continuous function and has acontinuous right inverse, i.e.g: YXsuch thatfg= idY, thenfis a an identificationmap. Proof: The existence ofg guarantees the surjectivity off. LetUbe a subset ofY.

Suppose f1(U) is open in X. Then U=id1Y (U) = (f g)1(U) =g1(f1(U)) is opendue to the continuity ofg. So fis an identification map.

Consider the map : R2 R, (x, y) = x. Define A R2 by A ={(x, y)|y =0 or x 0}, and define f = |A. Thenf is clearly continuous, and has an (obviouslycontinuous) right inverse defined byg : R R2,g(x) = (x, 0). By the lemma above, f isan identification map.

Let the graph ofy = 1/x be G, a closed subset ofA. f(G) = (0,) is not closed inR. Sof is not a closed map.

Consider the subset B = ((0, 2), 1) A in S. Since ((0, 2), 1) is open in R, B is openin A. f(B) = [0, 1) is not open in R. Sof is not open.


Solution:

It is easy to see thatf(x,y,z) = (x2y2,xy,xz,yz)is a (continuous) map. Since S2 iscompact and E4 is Hausdorff,fis a closed map. Since the projective plane can be obtained

by identifying antipodal points on S2 and a one-one closed map is an embedding, if we

can show that f(x,y,z) = f(x, y, z) either(x,y,z) = (x, y, z) or (x,y,z) =(

x,

y,

z), thenfis an embedding.

Clearly the map is well-defined (f(x,y,z) =f(x,y,z)), and f(x,y,z) =f(x,y, z).Suppose f(x,y,z) = (A,B,C,D) = f(x, y, z). Thenx2 y2 = A = x2 y2 and

xy= B = xy. Solving forx and y, we obtainx =(A+ A2 + 4B2)/2,y =2B/(A +A2 + 4B2). Since(x, y)is interchangeable with(x, y)in the solving procedure, we must

conclude that either (x, y) = (x, y) or (x, y) = (x,y). Plugging inx into xz = Cand then y into yz = D, we see that either either(x,y,z) = (x, y, z) or (x,y,z) =

(x,y,z).


Solution:

A natural way to construct a Klein bottle from[0, 2] [0, ]is identifying {0} [0, ]with{2} [0, ]in a parallel manner, and [0, 2]{0}with[0, 2]{}in a antiparallel

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manner. We will show that this indeed is the construction here. To do so, using the same

argument as in problem 10, we only need to show that f(x, y) =f(x, y) either (i)(x, y) = (x, y), or (ii) x, x {0, 2} and y =y , or (iii) y, y {0, } and x = 2 x(or x = x = 0 or x = x = 2, to account for the identification of all four corners of

[0, 2] [0, ]).It is easy to see that the map is well-defined (f(x, y) = f(x, y)), f(0, y) = f(2, y),

and f(x, 0) =f(2 x, ).Suppose f(x, y) = (A,B,C,D,E) = f(x, y). First, consider the case y= y. The

equations cos2y = B and sin2y = C together determine a unique value of 2y on the

parametrized circle, so ify=y , then clearly, say,y = 0and y =. Sof(x, 0) =f(x, ).Thencos x= A = cosx andsin x= D = sin x. Notice thatx= x = 0andx= x = 2satisfy these equations. So suppose x, x (0, 2). Clearly only the relation x = 2 xsatisfies the two trigonometric equations.

Now we consider the case y = y

(and x= x

): f(x, y) = f(x

, y). From cosx =A = cos x and sin x sin y = E = sin x sin y, which determine a unique point x on the

parametrized circle, the only satisfying relation isx, x {0, 2}.Sofinduces an embedding of the Klein bottle.


Solutions:

Suppose

(2 + cos x)cos2y= (2 + cos x)cos2y (1)

(2 + cos x)sin2y= (2 + cos x)sin2y (2)

Consider the case y, y / {/4, 3/4}. Thencos2y= 0= cos2y. Dividing (2) by(1) yields tan2y = tan2y. So either y = y or y = y +/2. If y = y +/2, then

cos2y = cos2y, so (1) yields 4 = cosx cos x, which is impossible. Thereforey= y . Socos 2y= cos 2y, sin 2y= sin 2y, and from (1) cos x= cos x.

Consider the case{y, y} {/4, 3/4} =. Suppose y = /4. Then (1) yields

cos2y

= 0, i.e. eithery

=y = /4or y

= 3/4. Ify

= 3/4, thensin 2y

= sin2y, so(2) yields 4 = cosx cosx, which is impossible. So y =y = /4. So cos 2y= cos 2y,sin2y= sin 2y, and from (1)cos x= cosx. Similar argument for the casey = 3/4leads

to the same argument.

Therefore, equations (1) and (2) necessarily yields cos2y = cos2y, sin2y = sin2y,

and cos x= cosx

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Using the result above, we see thatg(x, y) =g(x, y) f(x, y) =f(x, y). There-fore, we can use the same argument used in problem 11 to show that g is an embedding

of the Klein bottle in E4.

6 Example 1.3, lecture noteProblem:

Define the equivalence relationon R as xyx y Q. Show that R/is anuncountable set with indiscrete topology (and therefore is not Hausdorff).

Solution:

We first prove that R/is an uncountable set. It is clear that each equivalence classhas countably many representatives: for each equivalence class [r] with r R, we candefine a surjective mapg : Q [r]byg(q) =r if there exists r [r]such thatr r= qand g(q) = r otherwise. Then, ifR/ is a countable set, then R, being a countableunion of countable sets, must be countable (ridiculous).

Let U be an open, non-empty subset ofR/. We will show thatU = R/. Thekey of the solution is that the preimage of any non-empty open subset ofR/, by thedense-ness ofQ in R, lacks no representative of any equivalence classes on R, and therefore

must be mapped to the whole quotient space.

Let

f : R R/x[x]

be the canonical projection. SinceUis open, non-empty and sincefis continuous,f1(U)

is open and non-empty in R. Clearly it must contain some open interval I ofR. Let x

be any real number. Since Qis dense in R, we can find some q Q such thatx q= xfor some x I. Since (x x)Q, f(x) = [x] = [x] =f(x)f(I)f(f1(U))U.ThereforeR/=f(R)U, and soU= R/.

7 Example 1.12, lecture note

Problem: Prove that:

(a)Sn=Dn/Sn1(b)(Sn I)/(Sn {0})=Dn+1

Solutions:

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(a) Let : Dn Dn/Sn1 be the canonical projection. For a moment, suppose wehave a surjective (continuous) map f : Dn Sn such that it maps all ofSn1 to somep Sn and is bijective on Dn \ Sn1 (mapped to Sn \ {p}). Next, define the followingmap:

h: Dn

/Sn1

Sn

xf(1(x))

. hthen is clearly bijective. Since is an identification map and f=h is continuous,h is continuous by proposition 1.6 (l.n.). SinceDn is compact and Sn is Hausdorff, by

corollary 1.9 (l.n.), h is a homeomorphism.

We construct the map f as follows. Since (Dn \Sn1)= En = (Sn {p}), wehave a homeomorphism k :Dn \ Sn1 Sn {p}. Then we can definef(x) =k(x) forx

Dn

\Sn1 andf(x) =pforx

Sn1. The continuity offcan be shown as follows. Let

Ube an open proper subset ofSn containingp. ThenU\{p}= En\Dn=(Dn\Sn1)\C,whereC Dn \ Sn1 is closed. Noting that any open neighborhood ofSn1 must be ofthe form Dn \ C, it is easy to see that f1(U) is of such form and open. Then f is thedesired map.

(b) We can use the same technique in part (a), since (Sn I), being a closed andbounded subset ofEn+2, is compact and Dn+1 is Hausdorff. Note thatDn+1 \ {0}=Sn I\Sn {0} (by identifying points on Dn+1 = Sn with (t, 1) Sn {1}, andall points, say p, on int(Dn+1 \ O) with (t, s) Sn (I\ {1}), where s is the smallest

Euclidean distance from p to points on Dn+1

); let such a homeomorphism be k. Theneeded map fromSnItoDn+1 can be constructed asf(x) =k(x)forxSn (I\{0})andf(x) ={0}for xSn {0}. The rest of the proof is exactly the same as in part (a).

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