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.

  • 1
    $\begingroup$ It may be circumventing the intent of the exercise, but couldn't you just argue that your map is the "same" as the map in the long exact sequence, so induces an isomorphism? $\endgroup$ May 1, 2012 at 14:22
  • $\begingroup$ @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? $\endgroup$ May 1, 2012 at 14:28
  • $\begingroup$ @probably123 There is no isomorphism on unreduced homology (Think about $X=S^0$ and $\Sigma X = S^1$), so it makes sense to use the reduced chain complex. $\endgroup$ Dec 2, 2020 at 15:30
  • $\begingroup$ @JustinYoung You're right but I was just wondering that someone can give a proof using the map defined by the questioner. Also, in Hatcher, reduced homology is defined by the sequence $\cdots C_n(X) \to C_{n-1}(X) \to \cdots \to C_0(X) \to \Bbb Z\to 0$ which is obtained from the original complex by adjoining the map $C_0(X)\to \Bbb Z$, so it makes sense to define a chain map $C_n(X)\to C_{n+1}(SX)$. $\endgroup$
    – blancket
    Dec 2, 2020 at 15:51
  • $\begingroup$ It would have to be the augmented complex you wrote down (there is a group in dimension "-1"). You should try to make the map in the question work, you just need to do a bit of calculation, or prove it is homotopic (up to sign) to the map I defined (considered on the augmented complex), by constructing a chain homotopy. $\endgroup$ Dec 2, 2020 at 20:00

2 Answers 2


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.

  • $\begingroup$ Thank you for your answer. There's something I do not know (it is not in Hatcher's section prior to the exercise): what is the long exact homology sequence of a triple? $\endgroup$ May 2, 2012 at 18:20
  • 1
    $\begingroup$ Ah, well say you have a triple of subspaces $A\subset B \subset Y$, then if you think through the algebra, this induces a short exact sequence of abelian groups: $0\to C_*(B, A) \to C_*(Y, A) \to C_*(Y,B) \to 0$, the induced long exact sequence on homology is the exact sequence of a triple. In the example above, $Y = CX$, $B = X$, and $A = *$. $\endgroup$ May 2, 2012 at 18:29
  • $\begingroup$ Ah, I see. Thanks, I'll think this through a little later. $\endgroup$ May 2, 2012 at 18:47
  • $\begingroup$ @JustinYoung, I am stuck at this problem for some time now and stumbled upon your solution. One thing I cannot understand is that to apply long exact sequence, you need the maps in the chain groups to form a short exact sequence. I cannot see how the maps defined here is exact. $\endgroup$
    – ChesterX
    Jan 7, 2015 at 8:36
  • $\begingroup$ It's basically one of the isomorphism theorems $G/H \cong (G/N)/(H/N)$. $\endgroup$ Feb 22, 2016 at 16:20

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 $\sigma$.


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