Retract of a quasi-isomorphism of chain complexes.

Question

$\require{AMScd}$

I want to show that the retract of a quasi-isomorphism of chain complexes is also a quasi-isomorphism of chain complexes. Let $X,Y,A,B$ be chain complexes in degrees greater than and equal to $0$. Say that $f:X\to Y$ is a qis, and $g:A\to B$ is a retract of $f$:

$$\begin{CD}
A@>>> X@>>>A\\
@VVV @VVV @VVV\\
B @>>> Y@>>> B
\end{CD}$$
where the top composite is the identity, the bottom composite is the identity, $X\to Y$ is a qis and the diagram commutes.

Then taking homology and considering any degree $n$ we have:
$$\begin{CD}
H_n(A)@>>> H_n(X)@>>>H_n(A)\\
@VVV @VVV @VVV\\
H_n(B) @>>> H_n(Y)@>>> H_n(B)
\end{CD}$$

• The top row is the identity composition,
• Thus $H_n(A)\to H_n(X)$ is injective,
• Then the composite $H_n(A)\to H_n(B)\to H_n(Y)$ is injective,
• Thus the left square has $H_n(A)\to H_n(B)$ injective.
• Similarly $H_n(Y)\to H_n(B)$ is surjective.
• Since the right square commutes we have that $H_n(A)\to H_n(B)$ is surjective.
• But the injective map on the left column is the same map $g_*$ as the surjective map on the rightmost column.
• Then $H_n(A)\to H_n(B)$ is an isomorphism.
• This occurs in each degree and hence $A\to B$ is a qis.

Is this correct?

Answer 1

This still looks correct to me.