Quasi isomorphisms are weak equivalences for the category of chain complexes

Question

I am having a bit of difficulty proving that quasi isomorphisms are indeed weak equivalences for $\textbf{Ch(A)}$ (category of chain complexes) where A is any abelian category. The problem is in the gluing axiom.

The axiom states in this below mentioned diagram if $f$ and $f'$ are cofibrations (here in this case they are degree wise monomorpphism) and all the downward vertical arrows are weak equivalences (here quasi isomorphism).

$\require{AMScd}$ \begin{CD} C @<{g}<< A @>{f}>> B\\ @V{\sim}VV @V{\sim}VV @V{\sim}VV\\ C' @<{g'}<< A'@>{f'}>>B' \end{CD}

Then the induced map $B\cup_{A}C \longrightarrow B'\cup_{A'}C'$ is also a weak equivalence Meaning we have to show the map is a quasi isomorphism.

My attempt
I have to prove that $H_{n}(B.\cup_{A.}C.) \cong H_{n}(B.'\cup_{A.'}C.') $

Now if I follow the diagram then I have two exact sequences $$ 0 \longrightarrow C \longrightarrow B.\cup_{A.}C. \longrightarrow B/A \longrightarrow 0$$

and

$$ 0 \longrightarrow C.' \longrightarrow B'.\cup_{A'.}C'. \longrightarrow B'/A' \longrightarrow 0$$

So I was trying to use long exact sequence property of the homology, but it seems that I am stuck. Since pushout of objects might not be pushout of homology I also think I can not write $H_{n}(B.\cup_{A.}C.) \cong H_{n}(B.)\cup_{H_{n}(A.)}H_{n}(C.)$

It seems I am missing something very obviuos any hint or help is really appreciated. thanks in advance.

Answer 1

Your short exact sequences are related by maps that fit into a beautiful commutative diagram :

$$\require{AMScd}\begin{CD} 0@>>> C @>>> B\cup_A C @>>> B/A @>>> 0 \\ @VVV @VVV @VVV @VVV @VVV \\ 0@>>> C' @>>> B'\cup_{A'}C' @>>> B'/A' @>>> 0\end{CD}$$

therefore there is also a map of long exact sequences :

$$\require{AMScd}\begin{CD} H_{n+1}(B/A)@>>> H_n(C) @>>> H_n(B\cup_A C) @>>> H_n(B/A) @>>> H_{n-1}(C) \\ @VVV @VVV @VVV @VVV @VVV \\ H_{n+1}(B'/A')@>>> H_n(C') @>>> H_n(B'\cup_{A'} C') @>>> H_n(B'/A') @>>> H_{n-1}(C') \end{CD}$$

By the $5$-lemma, as $H_*(C)\to H_*(C')$ is an isomorphism, it suffices to show that the same holds for $H_*(B/A)\to H_*(B'/A')$ (and in fact, still by the $5$-lemma, it's equivaent to that requirement)

This is where the cofibration hypothesis will be useful (Note : the hypothesis essentially says that $0\to A \to B \to B/A \to 0$ is a short exact sequence).Is that enough for you to conclude ?

Answer 2

To conclude on Max's wonderful answer the below mentioned exact sequences forms a commutative diagram so now we use the long exact sequence of homology and five lemma to conclude $H_n(B/A) \cong H_n(B'/A')$

$\require{AMScd}$ \begin{CD} A @>>> B @>>> B/A\\ @V{\sim}VV @V{\sim}VV @VVV\\ A' @>>> B'@>>>B'/A' \end{CD}

So I will have a long exact sequence of homology

$\require{AMScd}$ \begin{CD} H_{n}(A) @>>> H_{n}(B) @>>> H_n(B/A) @>>> H_{n-1}(A)@>>>H_{n-1}(B)\\ @V{\cong}VV @V{\cong}VV @VVV @V{\cong}VV @V{\cong}VV\\ H_{n}(A') @>>> H_{n}(B') @>>> H_n(B'/A') @>>> H_{n-1}(A')@>>>H_{n-1}(B') \end{CD}

So by 5 lemma middle one is isomorphism and we can do it for each n thus the required isomorphism.