Explicit isomorphism $\tilde{H_n}(X)\simeq \tilde{H}_{n+1}(SX)$.

Question

Let $X$ be a topological space and $S:\mathbf{Top}\to \mathbf{Top}$ be the suspension functor.

It's not hard to show using e.g. the long exact sequence of homology that $\tilde{H_n}(X)\simeq \tilde{H}_{n+1}(SX)$ (where $\tilde{H_n}$ denotes reduced homology).

However, I need to

"construct explicit chain maps $f:C_n(X)\to C_{n+1}(SX)$ inducing isomorphisms $\tilde{H_n}(X)\to \tilde{H_{n+1}}(SX)$"

(this is problem 21 in section 2.1 of Hatcher's book).

Here's my attempt:

Define $f:C_n(X)\to C_{n+1}(SX)$ as follows. If $\sigma:\Delta^n\to X$ is a singular $n$-chain, then its suspension is $S\sigma:S\Delta^n\to SX$.

It's geometrically clear that $S\Delta^n$ is the union of two $n+1$- standard simplexes, call them $\Delta^{n+1}_0$ and $\Delta^{n+1}_1$, identified by a face.

Let $f(\sigma):=S\sigma|_{\Delta^{n+1}_1}-S\sigma|_{\Delta^{n+1}_0}$. Then extend $f$ linearly to all of $C_n(X)$.

A little manipulation proves that $f$ is indeed a chain map.

Now the problem is: how to prove that $f_*:\tilde{H_n}(X)\to \tilde{H}_{n+1}(SX)$ is an isomorphism?

I thought perhaps the connecting homomorphism $\partial:H_{n+1}(CX,X)\to \tilde{H}_n(X)$ could help. Since $(CX,X)$ is a good pair, there is an isomorphism $\varphi:\tilde{H_{n+1}}(CX/X)=\tilde{H}_{n+1}(SX)\to H_{n+1}(CX,X)$.

But how to prove that $\partial \varphi$ is the inverse of $f_*$?

Or perhaps this is a terrible approach... but I've run out of ideas.

Answer 1

I believe your map will work, but here's my suggestion.

We work with relative groups everywhere, since we want the conclusion to hold for reduced homology.

Consider the maps

$$C_*(X, *) \to C_{*+1}(CX, X) \to C_{*+1}(SX, *)$$

The first map is defined by taking a simplex $\Delta^n \to X$ to the simplex $\Delta^{n+1} = C\Delta^n \to CX$. The second map is the collapse map $(CX, X) \to (CX/X, X/X) = (SX, *)$. The second map is automatically a chain map, and it should be easy to check that the first map is also a chain map.

Consider the long exact homology sequence of the triple $(CX, X, *)$, then it should be easy to see, just by computing the map directly following chains around, that the connecting map $\delta: H_{*+1}(CX, X) \to H_*(X, *)$ is inverse to the induced map on homology $H_*(X, *) \to H_{*+1}(CX, X)$.

Using excision or what have you, you can then prove that the map $C_{*+1}(CX, X) \to C_{*+1}(SX, *)$ induces an isomorphism on homology. This shows that the above composition does the job.

I think if you do some work you can prove this map is homotopic to the map that you constructed, but this seems easier to me.

Answer 2

There is actually a way to make OP's approach work directly by using the Mayer–Vietoris sequence. The argument is more or less analogous to the other answer.

We can view $SX$ as the union of two cones $CX$ whose intersection is a neighborhood that deformation retracts to $X$. Observe that the boundary map $\tilde{H}_{n + 1}(SX) \to \tilde{H}_n(X)$ in the associated Mayer–Vietoris sequence is an isomorphism since the $\tilde{H}_*(CX) \oplus \tilde{H}_*(CX)$ terms all vanish.

Let us understand how the boundary map $\tilde{H}_{n + 1}(SX) \to \tilde{H}_n(X)$ works. This just involves working through the proof of the Mayer–Vietoris sequence. Consider an $(n + 1)$-cycle $\alpha$ in $SX$. By subdivision, $\alpha$ is homologous to a cycle $\beta - \gamma$ for a chain $\beta$ in the first $CX$ and a chain $\gamma$ in the second $CX$. Since $\alpha$ is a chain, $\beta$ and $\gamma$ have the same boundary, which is contained in the intersection of the two cones. Therefore, the boundary map $\partial$ is defined by $\partial[\alpha] = [\partial \beta] = [\partial \gamma]$.

Now, it should be clear that the chain map defined in the original post is an inverse of the boundary map. This is because $f\sigma$ can be subdivided into the two chains $S\sigma|_{\Delta_1^{n + 1}}$ and $S\sigma|_{\Delta_2^{n + 1}}$ (using notation from the original post) each of whose boundary is $\sigma$.