Fall 2017 Final

1

Let X be the subspace of the unit cube I3 consisting of the union of the 6 faces and the 4 internal diagonals. Compute π1(X).

2

Let X be an arbitrary topological space, and compute π1(ΣX).

solution:

Write ΣX=UV where U=ΣX(X×[0,1/2]) and U=ΣXX×[1/2,1]). Then UV=X×{1/2}X, so π1(UV)=π1(X).

But both U and V can be identified by the cone on X, given by CX=X×IX×1, by just rescaling the interval with the maps:

iU:UCX where (x,s)(x,2s1) (The second component just maps [1/2,1][0,1]. )

iV:VCX where (x,s)(x,2s). (The second component just maps [0,1/2][0,1])

But CX is contractible by the homotopy H:CX×ICX where H((c,s),t)=(c,s(1t)).

So π1(U)=π1(V)=0.

By Van Kampen, we have π1(X)=0π1(X)0=0.

3

Let X=S1×S1 and AX be a subspace with AS1S1. Show that there is no retraction from X to A.

solution:

We have π1(S1×S1)=π1(S1)×π1(S1) since S1 is path-connected (by a lemma from the problem sets), and this equals Z×Z.

We also have π1(S1S1)=π1(S1){pt}π1(S1), which by Van-Kampen is ZZ.

Suppose X retracts onto A, we can then look at the inclusion ι:AX. The induced homomorphism ι:π1(A)π1(X) is then also injective, so we’ve produced an injection from f:ZZZ×Z.

This is a contradiction, because no such injection can exists. In particular, the commutator [a,b] is nontrivial in the source. But f(aba1b1)=f(a)f(b)f(a)1f(b)1 since f is a homomorphism, but since the target is a commutative group, this has to equal f(a)f(a)1f(b)f(b)1=e. So there is a non-trivial element in the kernel of f, and f can not be injective - a contradiction.

4

Show that for every map f:S2S1, there is a point xS2 such that f(x)=f(x).

solution:

Suppose towards a contradiction that f does not possess this property, so there is no xS2 such that f(x)=f(x).

Then define g:S2S1 by g(x)=f(x)f(x); by assumption, this is a nontrivial map, i.e. g(x)0 for any xS2.

In particular, g(x)=(f(x)f(x))=f(x)f(x)=g(x), so g(x)=g(x) and thus g commutes with the antipodal map α:S2S2.

This means g is constant on the fibers of the quotient map p:S2RP2, and thus descends to a well defined map ˜g:RP2S1, and since S1RP1, we can identify this with a map ˜g:RP2RP1 which thus induces a homomorphism ˜g:π1(RP2)π1(RP1).

Since g was nontrivial, ˜g is nontrivial, and by functoriality of π1, ˜g is nontrivial.

But π1(RP2)=Z2 and π1(RP1)=Z, and ˜g:Z2Z can only be the trivial homomorphism - a contradiction.

remark:

Alternate Solution Use covering space R?

5

How many path-connected 2-fold covering spaces does S^1 \vee {\mathbf{RP}}2 have? What are the total spaces?

solution:

First note that \pi_1(X) = \pi_1(S^1) \ast_{{\operatorname{pt}}} \pi_1({\mathbf{RP}}2) by Van-Kampen, and this is equal to {\mathbf{Z}}\ast {\mathbf{Z}}_2.

6

Let G = <a, b> and H \leq G where H = <aba^{-1}b^{-1},~ a^2ba^{-2}b^{-1},~ a^{-1}bab^{-1},~ aba^{-2}b^{-1}a>. To what well-known group is H isomorphic?