Let \(X\) be the subspace of the unit cube \(I^3\) consisting of the union of the 6 faces and the 4 internal diagonals. Compute \(\pi_1(X)\).

Solution:

2

Let \(X\) be an arbitrary topological space, and compute \(\pi_1(\Sigma X)\).

Solution:

Write \(\Sigma X = U \cup V\) where \(U = \Sigma X - (X\times[0,1/2])\) and \(U = \Sigma X - X\times[1/2, 1])\). Then \(U\cap V = X \times\{1/2\} \cong X\), so \(\pi_1(U\cap V) =\pi_1(X)\).

But both \(U\) and \(V\) can be identified by the cone on \(X\), given by \(CX = \frac{X \times I}{X \times 1}\), by just rescaling the interval with the maps:

\(i_U: U \to CX\) where \((x,s) \mapsto (x, 2s-1)\) (The second component just maps \([1/2, 1] \to[0,1]\). )

\(i_V: V \to CX\) where \((x, s) \mapsto (x, 2s)\). (The second component just maps \([0,1/2] \to [0, 1]\))

But \(CX\) is contractible by the homotopy \(H:CX \times I \to CX\) where \(H((c,s), t) = (c, s(1-t))\).

So \(\pi_1(U) = \pi_1(V) = 0​\).

By Van Kampen, we have \(\pi_1(X) = 0 \ast_{\pi_1(X)} 0 = 0.\)

3

Let \(X = S^1 \times S^1\) and \(A\subset X\) be a subspace with \(A \cong S^1 \vee S^1\). Show that there is no retraction from \(X\) to \(A\).

Solution:

We have \(\pi_1(S^1 \times S^1) = \pi_1(S^1) \times\pi_1(S^1)\) since \(S^1\) is path-connected (by a lemma from the problem sets), and this equals \({\mathbf{Z}}\times{\mathbf{Z}}\).

We also have \(\pi_1(S^1 \vee S^1) = \pi_1(S^1) \ast_{\left\{{pt}\right\}} \pi_1(S^1)\), which by Van-Kampen is \({\mathbf{Z}}\ast {\mathbf{Z}}\).

Suppose \(X\) retracts onto \(A\), we can then look at the inclusion \(\iota: A \hookrightarrow X\). The induced homomorphism \(\iota_*: \pi_1(A) \hookrightarrow\pi_1(X)\) is then also injective, so we’ve produced an injection from \(f: {\mathbf{Z}}\ast {\mathbf{Z}}\hookrightarrow{\mathbf{Z}}\times{\mathbf{Z}}\).

This is a contradiction, because no such injection can exists. In particular, the commutator \([a,b]\) is nontrivial in the source. But \(f(aba^{-1}b^{-1}) = f(a)f(b)f(a)^{-1}f(b)^{-1}\) since \(f\) is a homomorphism, but since the target is a commutative group, this has to equal \(f(a)f(a)^{-1} f(b)f(b)^{-1} = e\). 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 \(f: S^2 \to S^1\), there is a point \(x\in S^2\) such that \(f(x) = f(-x)\).

Solution:

Suppose towards a contradiction that \(f\) does not possess this property, so there is no \(x\in S^2\) such that \(f(x) = f(-x)\).

Then define \(g: S^2 \to S^1\) by \(g(x) = {f(x) - f(-x)}\); by assumption, this is a nontrivial map, i.e. \(g(x) \neq 0\) for any \(x\in S^2\).

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 \(\alpha: S^2 \to S^2\).

This means \(g\) is constant on the fibers of the quotient map \(p: S^2 \to{\mathbf{RP}}2\), and thus descends to a well defined map \(\tilde g: {\mathbf{RP}}2 \to S^1\), and since \(S^1 \cong {\mathbf{RP}}1\), we can identify this with a map \(\tilde g: {\mathbf{RP}}2 \to{\mathbf{RP}}1\) which thus induces a homomorphism \(\tilde g_*: \pi_1({\mathbf{RP}}2) \to \pi_1({\mathbf{RP}}1)\).

Since \(g\) was nontrivial, \(\tilde g\) is nontrivial, and by functoriality of \(\pi_1\), \(\tilde g_*\) is nontrivial.

But \(\pi_1({\mathbf{RP}}2) = {\mathbf{Z}}_2\) and \(\pi_1({\mathbf{RP}}1) = {\mathbf{Z}}\), and \(\tilde g_*: {\mathbf{Z}}^2 \to{\mathbf{Z}}\) can only be the trivial homomorphism - a contradiction.

Alternate Solution

Use covering space \({\mathbf{R}}\twoheadrightarrow S^1\)?

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?

Solution: