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Suppose I have a topological space $X$, and its cone $CX = ([0,1] \times X)/(\{0\} \times X)$. I have a map $s : C_n(X) \to C_{n+1}(CX,X)$ given by sending singular $n$-simplices to the “natural” $(n+1)$-simplices obtained by identifying $\Delta^{n+1} = C\Delta^n$ (see the answer to this question for an explicit definition of $s$).

I want to show: $s : C_n(X) \to C_{n+1}(CX,X)$ is a chain map inducing an isomorphism $H_n(X) \cong H_{n+1}(CX,X)$. As I understand it, this means I need to show $\partial s = s\partial$. However, I have instead found that $s$ anti-commutes with the boundary: $\partial s = -s\partial$.

That being said, this means $s$ takes cycles to relative cycles and boundaries to relative boundaries, so the induced homomorphism $s_* : H_n(X) \to H_{n+1}(CX,X)$ is well-defined, and is an inverse to the connecting homomorphism $H_{n+1}(CX,X) \to H_n(X)$ (which is in fact an isomorphism, by the corresponding long exact sequence since $CX$ is contractible, so $s_*$ is also an isomorphism).

Modulo the fact that $s$ doesn’t seem to be a chain map, I’ve solved the problem. But in some comments to the linked question, it’s mentioned that a “skewed” chain map needs to anti-commute with the boundary, not commute.

Why does a chain map $s : C_n(X) \to C_{n+1}(CX,X)$ need to anti-commute with the boundary? If we think of $C_n(X) = A_n$ and $C_{n+1}(CX,X) = B_n$ as abstract abelian groups and associated chain complexes, why would a chain map not simply commute with the boundary? The fact that our chain complexes come from topological spaces seems like superfluous information in the context of defining a chain map.

(I’m including the “homological algebra” tag because I think my confusion is about what exactly a “chain map” abstractly is; if that’s not an appropriate tag and this is too trivial for real homological algebra, I apologize.)

D Ford
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    This is discussed in the comments of the answer you link to. Have you checked the link provided there? Also check out this post. – Thorgott Oct 20 '21 at 18:23
  • It’s discussed, yes, but not justified beyond “that’s just how it’s done”. I’m looking for either some intuition or why it must be defined in this way (Hatcher, for example, only defines chain maps when the groups are “aligned”, so I only know how to define chain maps for general chain complexes). The other post provides a little more justification, I think, but still seems to assume more algebraic structure (graded algebras in particular) than I’d like to invoke if possible (since the setting of abelian groups in chain complexes is pretty simple and natural) – D Ford Oct 20 '21 at 18:49

1 Answers1

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Your map actually makes the diagram commute, it's just that you're shifting a complex, so its differential changes. In short, you have a complex $(C,d)$ corresponding to $C_*(X)$, and you have a complex $(D,d)$ corresponding to $C_*(CX)$. But what you are giving is a map from $(C,d)$ to the desuspension $s^{-1}D$ of $(D,d)$. This is a complex that satisfies $(s^{-1}D)_n = D_{n+1}$ and whose differential is computed by the Koszul sign rule:

$$ d(s^{-1}x) = (-1)^{|s^{-1}||d|} s^{-1} dx = -s^{-1} dx$$.

So the differential in the suspension $C_{*+1}(CX)$ is $-d$, and not $d$.

Add. Please do not confuse the desuspension sign $s^{-1}$ above with your map!

Pedro
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