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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.

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    @Jim: By "your map", do you mean $\partial \varphi$? Yes, being the composition of two isomorphisms it is an isomorphism, but how could I link it to $f_*$? Or maybe I'm misunderstanding your comment, did you mean something else?2012-05-01

1 Answers 1

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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.