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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.

Bruno Stonek
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    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? – Cheerful Parsnip May 01 '12 at 14:22
  • @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? – Bruno Stonek May 01 '12 at 14:28
  • @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. – Justin Young Dec 02 '20 at 15:30
  • @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 $C_n(X)\to C_{n+1}(SX)$. – blancket Dec 02 '20 at 15:51
  • 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. – Justin Young Dec 02 '20 at 20:00

2 Answers2

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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.

Justin Young
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  • 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? – Bruno Stonek May 02 '12 at 18:20
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    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 = $. – Justin Young May 02 '12 at 18:29
  • Ah, I see. Thanks, I'll think this through a little later. – Bruno Stonek May 02 '12 at 18:47
  • @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. – ChesterX Jan 07 '15 at 08:36
  • It's basically one of the isomorphism theorems $G/H \cong (G/N)/(H/N)$. – Justin Young Feb 22 '16 at 16:20
  • @JustinYoung You can't apply the excision theorem to $C_{\ast + 1}(CX,X) \to C_{\ast +1}(SX,\ast)$, assuming $A,B \subset X$ are subspaces, $(B, A \cap B) \hookrightarrow (X,A)$ induce isomorphisms $H_n(B, A \cap B) \to H_n(X,A)$ for all $n$. In this notation, $A = \ast, B = CX, A \cap B = \ast$ and $X = SX$. I'm pretty sure that $X$ does not inject into $\ast$. Note that I've assumed $\ast$ denotes a point in $X$. –  Sep 28 '16 at 01:13
  • Not literally, but I did say "what have you"... Anyways, you can decompose the suspension into the top half and bottom half $SX = A\cup B$, then $A$ and $B$ are (homotopy equivalent to) $CX$, and $A\cap B$ is (homotopy equivalent to) $X$. Since $A$ and $B$ are (homotopy equivalent to) points, you get the result from excision modulo some homotopy equivalences. – Justin Young Sep 29 '16 at 19:40
  • @JustinYoung Modulo some homotopy equivalences? –  Sep 29 '16 at 23:31
  • Yes, the ones I mentioned in my comment. – Justin Young Sep 29 '16 at 23:56
  • I am able to prove that the connecting map is the left inverse which is equivalent as saying that it is surjective. But I am stuck on the other side. Can anyone give me some hints. – Arka Feb 04 '19 at 19:58
  • @ArkaGhosh The general theory (cone is contractible) tells you that the connecting map $\delta$ is an isomorphism. If you have an isomorphism and a one sided inverse, it must be a two sided inverse. – Justin Young Feb 04 '19 at 21:01
  • Why does the fact cone is contractible says that delta is an isomorphism. – Arka Feb 04 '19 at 21:06
  • @ArkaGhosh $H_(CX,) = 0$, so the connecting map is an isomorphism by exactness of the long exact sequence. – Justin Young Feb 04 '19 at 21:07
  • Oh yes thank you. Totally missed the trivial fact – Arka Feb 04 '19 at 21:12
  • @JustinYoung I have two questions. 1. I'm not sure that the map $C_* (X,) \to C_{+1}(CX,X)$ is well-defined. 2. Assuming that this map is well-defined, how do I have to show that this map is a chain map? – user302934 Sep 15 '19 at 12:30
  • The only issue with being well defined is the constant simplex at the base point, but that splits off $C_*(X)$ as a basis element, so we can just send that to zero. The map does work a bit better if we use the reduced cone, and then there is no need to define it separately. As for the chain map part, you need to look at the identification $\Delta^{n+1} \cong C\Delta^n$ and then just calculate the simplicial maps. – Justin Young Sep 15 '19 at 16:34
  • @Comol The details of the signs are worked out in my answer here: https://math.stackexchange.com/questions/3107083/the-naturality-of-the-cone-construction – Justin Young Sep 28 '19 at 11:42
  • @JustinYoung I'm not sure I see why the cone map $c: C_(X, ) \to C_*(CX, X)$ is a chain map. The other answer you've linked to shows that it is chain nullhomotopy of the identity, so $\partial c \equiv - c\partial$ modulo $X$, which is off by a negative factor. What am I missing? – abhi01nat Aug 29 '20 at 18:04
  • @abhi01nat The cone map has degree 1 (you are missing a "+1" in your degree for the codomain), so to be a chain map means it anti-commutes with the boundary. – Justin Young Aug 29 '20 at 18:37
  • I see. Do you have a reference for this? I have not been able to find a source for the definition of a chain map between chains that aren't "aligned". – abhi01nat Aug 30 '20 at 05:03
  • @abhi01nat It's the standard Koszul rule, I don't know a specific reference. Here's a MO post about it: https://mathoverflow.net/questions/354076/reason-to-apply-the-koszul-sign-rule-everywhere-in-graded-contexts – Justin Young Aug 31 '20 at 18:37
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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 $\sigma$.

Frank
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