Let $f:X\to Y$ be a map, and $\text{cone}(f) = CX \sqcup_f Y$ its mapping cone. Let $(H_n)_{n\in \Bbb{Z}}, (\partial_n)_{n\in \Bbb{Z}}$ be a homology theory with values in the category of $R$-modules. Then $H_n(f)$ is an isomorphism for all $n$ if and only if $H_n(\text{cone}(f), x_0) =0$ for all $n$ where $x_0$ is the cone point.

I tried using the Mayer-Vietoris sequence, applied to the cover of $\text{cone}(f)$ by $CX$ and $Y$. However, it didn't lead me anywhere since the homology theory is not required to satisfy the dimension axiom.

Thankful for any hint.

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    $\begingroup$ any restrictions on $X,Y$? $\endgroup$ – Daniel Valenzuela May 26 '15 at 21:02
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    $\begingroup$ $X,Y$ are any topological spaces $\endgroup$ – iwriteonbananas May 27 '15 at 5:35

I think the excision theorem on the triple $(\mbox{cone}(f), CX,\{x_0\})$ should help here. Then, the homology of $(\mbox{cone}(f),x_0)$ is isomorphic to the homology of $(X\times[0,1]\sqcup_f Y, X\times [0,1])$ (here we are gluing the end of the cylinder over $X$ to $Y$ by $(x,1)\sim f(x)$).

The exact sequence of this pair should get you the rest of the way because the inclusion map of $X\times [0,1]$ is essentially just $f$ now, and the two spaces deformation retract each onto $Y$ and $X$ respectively.

  • $\begingroup$ Sorry, could you expand on this a little bit? I'm not sure what you mean. Excision would tell us that $H_*((\text{cone}(f), CX) \cong H_*((\text{cone}(f) - x_0, CX - x_0)$. I don't see how the homology of $(\text{cone}(f), x_0)$ is involved. $\endgroup$ – Eric Auld Feb 22 '16 at 6:04
  • $\begingroup$ I'll admit I've forgotten my reasoning when I wrote this question, but it probably has something to do with the fact that $CX$ is contractible onto the point $x_0$, and so $(\operatorname{cone}(f),CX) \simeq (\operatorname{cone}(f),x_0)$. Here I guess we need to be careful with what we mean by $CX$. Let $CX=\{(x,t)\mid x \in X,\: t\in [0,\frac{1}{2}]\}$ so that we don't have this cone intersecting $Y$ and it really is homeomorphic to the usual cone over $X$. Then it's clear that $\operatorname{cone}(f)/CX \cong \operatorname{cone}(f)$. $\endgroup$ – Dan Rust Feb 22 '16 at 13:07

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