Consider two exact sequences of chain complexes (of abelian groups)
with morphisms of chain complexes $f, g, h$ between them:$\require{AMScd}$
$$
\begin{CD}
0 @>>> A_\bullet @>\alpha>> B_\bullet @>\beta>> C_\bullet @>>> 0\\
& @VfVV @VgVV @VhVV \\
0 @>>> X_\bullet @>\xi>> Y_\bullet @>\nu>> Z_\bullet @>>> 0
\label{d1} \tag{1}
\end{CD}
$$
If both squares in the diagram are commutative, the naturality of the zig-zag lemma gives us a morphism between the two long exact sequences, i. e. the diagram
$$
\begin{CD}
@>>> H_n(A_\bullet) @>>> H_n(B_\bullet) @>>> H_n(C_\bullet) @>\delta>> H_{n-1}(A_\bullet) @>>> \\
&@Vf_*VV @Vg_*VV @Vh_*VV @Vf_*VV \\
@>>> H_n(X_\bullet) @>>> H_n(Y_\bullet) @>>> H_n(Z_\bullet) @>\delta>> H_{n-1}(X_\bullet) @>>>
\end{CD}
$$
is commutative. Actually, the only nontrivial part of this is the commutativity of the the right square with the connecting morphisms $\delta$.
Question: What happens if the squares in diagram (\ref{d1}) are only commutative up to homotopy? My intuition told me that the zig-zag lemma should still be true, but I can't manage to prove it and I am actually starting to doubt whether this could really work. Again, the only nontrivial part is the commutativity of the squares with the $\delta$s.
Edit: I guess that this is equivalent to asking whether the zig-zag lemma is still valid in the homotopy category of chain complexes
Best Answer
This answer gives the following counterexample. $\newcommand\toby\xrightarrow$
Let $k$ be a field. Take $B= \cdots\to 0 \to k \toby{1} k \to 0 \to \cdots$, with $k$s in degrees $0$ and $1$. (Arrows going to the right for notational convenience, but still with homological grading.) Let $A$ be the subcomplex $\cdots \to 0 \to 0 \to k \to 0 \to \cdots$, with a $k$ in degree $0$. Then $C=B/A$ has a $k$ in degree $1$. Then the diagram $$ \require{AMScd} \begin{CD} 0 @>>> A @>>> B @>>> B/A @>>> 0 \\ @. @VV1V @VV1V @VV0V @. \\ 0 @>>> A @>>> B @>>> B/A @>>> 0 \\ \end{CD} $$ commutes up to homotopy. The first square clearly commutes, and $B$ is contractible, the nullhomotopy being $\cdots \leftarrow 0 \leftarrow k \overset{1}{\leftarrow} k \leftarrow 0 \leftarrow \cdots$, so any two maps out of $B$ are homotopic.
Then the LES in homology is the following $$ \begin{CD} 0=H_1(B) @>>> H_1(B/A) @>\delta>> H_0(A) @>>> H_0(B)= 0\\ @. @VV0V @VV1V @. \\ 0=H_1(B) @>>> H_1(B/A) @>\delta>> H_0(A) @>>> H_0(B) = 0,\\ \end{CD} $$ which becomes $$ \begin{CD} 0 @>>> k @>1>> k @>>> 0\\ @. @VV0V @VV1V @. \\ 0 @>>> k @>1>> k @>>> 0,\\ \end{CD} $$ which does not commute.