Let $f_*:C_*\to D_*$ be a chain map. I'm stuck in the proof of the following statement:
$f_*$ induces an isomorphism in homology iff the mapping cone of $f_*$, cone($f_*$), is contractible.
(For information about the mapping cone of a chain map see for example here https://ncatlab.org/nlab/show/mapping+cone , definition 6 and proposition 6).
Proof: Consider the short exact sequence of chain complexes
$$0\to D_*\to \operatorname{cone}(f_*)\to \Sigma C_*\to 0, $$where $i_*D_*\to \operatorname{cone}(f_*)$ is the inclusion and $p_*:\operatorname{cone}(f_*)\to \Sigma C_*$ is the projection onto the suspension of $C_*$.
Then there is a long exact sequence in singular homology
$$..\to H_n(D_*)\to H_n(\operatorname{cone}(f_*))\to H_n(\Sigma C_*)\xrightarrow{\partial_n} H_{n-1}(D_*)\to ..,$$furthermore it is $H_n(\Sigma C_*)\cong H_{n-1}( C_*)$. Now I'm stuck to prove that $\partial_n$ as a map $H_{n-1}( C_*)\to H_{n-1}( D_*)$ equals $H_{n-1}(f_*):H_{n-1}( C_*)\to H_{n-1}( D_*)$. I considered the explicit contruction of $\partial_*$ in the snake lemma, nevertheless I wasn't able to prove this. Can we elaborate it?
The next steps are: We have the long exact sequence $$..\to H_n(D_*)\xrightarrow{H_n(i_*)} H_n(\operatorname{cone}(f_*))\xrightarrow{H_n(p_*)} H_{n-1}( C_*)\xrightarrow{H_{n-1}(f_*)} H_{n-1}(D_*)\xrightarrow{H_{n-1}(i_*)} H_{n-1}(\operatorname{cone}(f_*))\to ..$$
$"\Leftarrow "$: If $\operatorname{cone}(f_*)$ is contractible, then $H_n(\operatorname{cone}(f_*))=0$ for all $n\in\mathbb{N}$. The sequence in homology is exact, it follows that $H_n(f_*)$ is an isomorphism for all $n\in\mathbb{N}$.
$"\Rightarrow "$: For fixed, but arbitrary $n\in\mathbb{N}$ the map $H_n(f_*)$ is an isomorphism. It follows: $\operatorname{ker}(H_{n+1}(p_*))=H_{n+1}(\operatorname{cone}(f_*))$ and $\operatorname{im}(H_{n+1}(i_*))=0$ ( because $H_{n+1}(f_*)$ is an isomorphism). But otherwise it is $\operatorname{im}(H_{n+1}(i_*))=\operatorname{ker}(H_{n+1}(p_*))$, i.e.: $H_{n+1}(\operatorname{cone}(f_*))=0$ and $H_{0}(\operatorname{cone}(f_*))=0$. It follows that $\operatorname{cone}(f_*)$ is contractible.
So the only thing which is not clear for me is that $\partial_n:H_{n-1}( C_*)\to H_{n-1}( D_*)$ equals $H_{n-1}(f_*):H_{n-1}( C_*)\to H_{n-1}( D_*)$. I appreciate your help.
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Best Answer
It's probably far easier (and instructive) to prove the fact directly. (Moreover it's possible for a chain complex to have vanishing homology but not be contractible, consider the chain complex $$\dots \to 0 \to \mathbb{Z} \xrightarrow{2 \cdot} \mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0$$ but the correct statement is indeed that $f_*$ is an isomorphism in homology iff the mapping cone has vanishing homology groups).
Recall that the mapping cone $Z_*$ is given by $Z_n = D_n \oplus C_{n-1}$ and $$d_Z(y,x) = (d_D(y) + f(x), -d_C(x)).$$
$(\implies)$ Suppose that $f_*$ is an isomorphism in homology. Let $[z] \in H_n(Z)$ be some homology class, represented by a cycle $z \in Z_n$. We want to show that $[z] = 0$, i.e. that $z$ is a boundary.
Write $z = (y,x) \in D_n \oplus X_{n-1}$. We know that $z$ is a cycle, i.e. $$d_Z(z) = 0 = (d(y) + f(x), -d(x)) \implies d(x) = 0 \text{ and } f(x) = d(-y).$$ Thus $x \in X_{n-1}$ is a cycle, and so we can consider $[x] \in H_{n-1}(C)$. But $f(x) = d(-y)$ means that $f(x)$ is a boundary, i.e. $f_*[x] = 0$. Since $f_*$ is an isomorphism, hence injective, and $[x] = 0$ too. So $x = d(x')$ is a boundary for some $x' \in C_n$ (Fact 1).
But now $0 = d(y) + f(x) = d(y) + f(d(x')) = d(y + f(x'))$, and so $y + f(x') \in D_n$ is a cycle, $[y + f(x')] \in H_n(D)$. But $f_*$ is an isomorphism, hence surjective, and thus $[y+f(x')] = f_*[x'']$ for some cycle $x'' \in C_n$, hence $y = f(x''-x')$ (Fact 2).
Combining the two facts, we get: $$d(0,x''-x') = (f(x''-x'), -d(x''-x') = (y,x) = z$$ as we wanted.
$(\impliedby)$ Now suppose that all the homology groups of the mapping cone vanish, i.e. $H_*(Z) = 0$. We want to show that $f_*$ is an isomorphism in homology.
$f_*$ is injective: Let $[x] \in H_n(C)$ be such that $f_*[x] = 0$, i.e. $dx = 0$ and $f(x) = dy$ for some $y \in D_{n+1}$. We want to show $[x] = 0$. But then $(y,-x) \in Z_{n+1}$ is a cycle: $d(y,-x) = (dy - f(x), -dx) = 0$. Since $H_{n+1}(Z) = 0$, it follows that $(y,-x)$ is a boundary, $(y,-x) = d(y',x') = (dy' + f(x'), -dx')$. In particular $x = dx'$ is a boundary, hence $[x] = 0$ as we wanted.
$f_*$ is surjective: let $[y] \in H_n(D)$ be represented by a cycle $y \in D_n$ ($dy = 0$). Then $(y,0) \in Z_n$ is a cycle, so it's a boundary (because $H_n(Z) = 0$): $(y,0) = d(y',x') = (d(y') + f(x'), -dx')$. So $x' \in C_n$ is a cycle, and $[y] = [d(y') + f(x')] = [f(x')] = f_*[x']$ is in the image of $f_* as we wanted.
I know it looks like a lot of words, but the underlying reasoning is not very complex. It's a simple diagram chase, and it mostly boils down to keeping track of all the definitions and hypotheses correctly. There's only one thing to do at each step, and so we do it (and the mapping cone is designed precisely so that everything works out).