Homology of Mapping Cone is Trivial--Proof from scratch

Question

Let $(A_*,\partial )$ and $(B_*,\partial' )$ be chain complexes, $f:A_*\to B_*$ a chain map and suppose that the map $f_*$ it induces on homology is an isomorphism.

The mapping cone $(C_*,D)$ is defined by $C_n=A_{n-1}\oplus B_n$ and $D_n(a_{n-1},b_n)=(-\partial_{n-1}a_{n-1},f_{n-1}(a_{n-1})+\partial'_n b_n).$

Now, define $A_n^+=A_{n-1}.$ Then, using the maps $i:b\mapsto (0,b)$ and $p:(a,b)\mapsto a$, there arises in the usual way, the long exact sequence

$$ \cdots \rightarrow H_{n+1}(C)\overset{p_*}\rightarrow H_{n+1}(A^+)\overset{f_*}\rightarrow H_n(B)\overset{i_*}\rightarrow H_n(C)\overset{p_*}\rightarrow H_n(A^+)\rightarrow \cdots $$

Then, $H_n(C)$ is trivial for all $n$. The proof is easy (but hadn't occurred to me, ugh.) I wanted to do it using the basic facts about the chain complexes involved, to get an idea of how they work.

There is a proof from scratch of essentially the same thing here but I think I found a problem with it, and made a comment there. I would appreciate feedback on my comment there as well as on my attempt here at a from scratch proof, which follows.

What I have so far (dropping the subscripts and primes on the differentiation operators):

Let $(a,b)\in Z_{n}(C).$ Then, by exactness, we have have the following:

$1).\ p_*([(a,b)]=[a]\Rightarrow a\in Bd_{n}A^+$ and $2).\ i_*([b])=[(0,b)]\Rightarrow (0,b)\in Bd_{n}C.$

So, there is an $a'\in A_n$ and a pair $(a'',b')\in C_{n+1}$ such that

$3).\ \partial a'=a$ and $4).\ (-\partial a'',f(a'')+\partial b')=(0,b).$

Combining these facts we get $(a,b)=(a,0)+(0,b)=(-\partial (a'+a''),f(a'')+\partial b').$

If I could show that either $f(a')=0$ or $\partial a'=0$, I would then have $(a,b)\in Bd_nC,$ which is what I want.

Answer 1

Assuming your question is why $C$ has trivial homology: $f_*$ is an isomorphism, so $p_*$ and $i_*$ must both be zero maps.

Answer 2

That is probably not the answer you are looking for, but here is a proof using model category theory. Somehow, it says that this is a result not about chain complexes per se, and it is a motivation not to dive into chain complexes details.

In any model category $\mathcal C$, the mapping cone $C_f$ of $f:X\to Y$ is defined as the homotopy pushout of $\ast \overset !\gets X \overset f\to Y$. Computing homotopy pushout in a model category goes as follow: take a cofibrant replacement $X'\overset q\to X$ of $X$. Factor $fq$ as $pi$ with $p$ acyclic fibration and $i$ cofibration. Factor $!q$ as $p'i'$ with $p'$ acyclic fibration and $i'$ cofibration. Then take the ordinary pushout of $\bullet \overset {i'} \gets X' \overset i \to \bullet$. This is summed up in the following diagram:

In particular, $q,p,p'$ are weak equivalences, so if $f$ is also a weak equivalence, we get that $i$ is an acyclic fibration. It follows that its pushout $j$ also is such. In the end, we have a commutative triangle:

where $p'$ and $j$ both are weak equivalences, which yields that $C_f \to \ast$ is also a weak equivalence. Otherwise put, $C_f$ is contractible.

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If you play that proof in your favorite model structure on chain complexes for which weak equivalences are the quasi-isomorphisms, you get the result you wanted.