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=U∪V where U=ΣX−(X×[0,1/2]) and U=ΣX−X×[1/2,1]). Then U∩V=X×{1/2}≅X, so π1(U∩V)=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:U→CX where (x,s)↦(x,2s−1) (The second component just maps [1/2,1]→[0,1]. )
iV:V→CX where (x,s)↦(x,2s). (The second component just maps [0,1/2]→[0,1])
But CX is contractible by the homotopy H:CX×I→CX where H((c,s),t)=(c,s(1−t)).
So π1(U)=π1(V)=0.
By Van Kampen, we have π1(X)=0∗π1(X)0=0.
3
Let X=S1×S1 and A⊂X be a subspace with A≅S1∨S1. 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(S1∨S1)=π1(S1)∗{pt}π1(S1), which by Van-Kampen is Z∗Z.
Suppose X retracts onto A, we can then look at the inclusion ι:A↪X. The induced homomorphism ι∗:π1(A)↪π1(X) is then also injective, so we’ve produced an injection from f:Z∗Z↪Z×Z.
This is a contradiction, because no such injection can exists. In particular, the commutator [a,b] is nontrivial in the source. But f(aba−1b−1)=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:S2→S1, there is a point x∈S2 such that f(x)=f(−x).
solution:
Suppose towards a contradiction that f does not possess this property, so there is no x∈S2 such that f(x)=f(−x).
Then define g:S2→S1 by g(x)=f(x)−f(−x); by assumption, this is a nontrivial map, i.e. g(x)≠0 for any x∈S2.
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:S2→S2.
This means g is constant on the fibers of the quotient map p:S2→RP2, and thus descends to a well defined map ˜g:RP2→S1, and since S1≅RP1, we can identify this with a map ˜g:RP2→RP1 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∗:Z2→Z can only be the trivial homomorphism - a contradiction.
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?