Quasi-isomorphisms are stable under homotopy base change in $\operatorname{Ch}(\mathcal{A})$ Edit: People are commenting that what I'm trying to do follows from "general theory of model categories" or "general theory of categories of fibrant objects". I have no idea at all of any of these theories and reading the definition from nLab does not tell me much. I only know a little of category theory and some theory of abelian categories. Here I give an elementary proof that only relies on abelian category theory. Any detailed explanation of how any of this exactly follows from more general abstract nonsense is very welcome, specially if it explains what is the model structure / fibrant objects structure on $\operatorname{Ch}(\mathcal{A})$ from necessary to make it work.
Before stating my question, I will explain the theoretical background. Let $\mathcal{A}$ be an additive category. In the category $\operatorname{CoCh}(\mathcal{A})$ of cochain complexes we can define an homotopy pullback in the following way: given morphisms $f:A\to C$ and $g:B\to C$ in $\operatorname{CoCh}(\mathcal{A})$, we define $A\times^h_CB$ as the cochain complex $(A\times^h_CB)^i=A^i\oplus B^i\oplus C^{i-1}$ with differential $d^i:(a,b,c)\mapsto(d^i_A(a),d^i_B(b),f^i(a)-g^i(b)-d^{i-1}_C(c))$. We have canonical projections
\begin{align*}
p&:A\times_C^hB\to A\\
q&:A\times_C^hB\to B
\end{align*}
such that the canonical projection $h^i:(A\times_C^hB)^{i}=A^{i}\oplus B^{i}\oplus C^{i-1}\to C^{i-1}$ defines the components of a cochain homotopy $h:fp\cong gq$. The object $A\times^h_CB$ is characterized by a universal property. Given a cochain complex $D$, define the set of triples
$$
\operatorname{Hom}(D,(f,g))
=\{(\tilde{f},\tilde{g},\tilde{h})\mid \tilde{f}:D\to A,\;\tilde{g}:D\to B
\text{ and } \tilde{h} \text{ is a cochain homotopy from } f\tilde{f}\text{ to }g\tilde{g}\}.
$$
Then $\operatorname{Hom}(-,(f,g))$ is a functor $\operatorname{CoCh}(\mathcal{A})^\mathrm{op}\to\mathsf{Set}$. This way, $A\times_C^hB$ is the cochain complex representing $\operatorname{Hom}(-,(f,g))$. That is, the map
\begin{align*}
\operatorname{Hom}_{\operatorname{CoCh}(\mathcal{A})}(D,A\times_C^hB)
&\to\operatorname{Hom}(D,(f,g))\\
\alpha&\mapsto(p\alpha,q\alpha,h\alpha)
\end{align*}
defines the components of a natural isomorphism of functors $\operatorname{CoCh}(\mathcal{A})^\mathrm{op}\to\mathsf{Set}$. Where $(h\alpha)^i=h^i\alpha^i:D^i\to C^{i-1}$.
I was trying to show the result "quasi-isomorphisms are stable under homotopy base change". That is, if in the diagram
$\require{AMScd}$
\begin{CD}
A\times^h_C B @>{p}>> A\\
@V{q}VV @VV{f}V\\
B @>>{g}> C
\end{CD}
(which commutes up to homotopy) we have that $g$ is a quasi-isomorphism (induces isomorphisms on all homology objects), then $p$ is a quasi-isomorphism. The following is the proof I was trying to construct:
We have a short exact sequence of chain complexes
$$
\tag{1}\label{1}
0\to C[-1]\to A\times_C^hB\xrightarrow{p\times q} A\oplus B\to 0,
$$ where the left map is the inclusion and the second one is the projection on the two first components. If we apply the result "short exact sequence of chain complexes induces long exact sequence on homology" to \eqref{1}, we obtain a LES
$$
\tag{2}\label{2}
\cdots \longrightarrow H^{i-1}(C) \longrightarrow H^{i}\left(A \times_{C}^{h} B\right) \xrightarrow{H^i(p)\times H^i(q)} H^{i}(A) \oplus H^{i}(B) \longrightarrow H^{i}(C) \longrightarrow \cdots
$$
Although I haven't worked out the details for an arbitrary additive category $\mathcal{A}$, for the category $\mathcal{A}=\mathsf{Ab}$ of abelian groups the boundary map $H^{i}(A) \oplus H^{i}(B) \to H^{i}(C) $ equals $H^i(f)-H^i(g)$. This is because in the commutative diagram of abelian groups with exact rows
$\require{AMScd}$
$$
\begin{CD}
@. C^{i-1}@>>> A^i\oplus B^i\oplus C^{i-1}@>>> A^i\oplus B^i@>>>0\\
@. @V{d^{i-1}_C}VV @V{d^i_{A\times_C^hB}}VV @V{d^i_A\oplus d^i_B}VV\\
0@>>> C^i@>>> A^{i+1}\oplus B^{i+1}\oplus C^{i}@>>> A^{i+1}\oplus B^{i+1}
\end{CD}
$$
we can do the chase

Now, if $H^i(g)$ is an isomorphism, in particular it is an epimorphism and thus the map $H^i(f)-H^i(g)=H^i(f)\amalg(-H^i(g))$ is also an epimorphism (follows from the identity $k(l\amalg m)=kl\amalg km)$ for the coproduct). This way, $H^{i-1}(C)\to H^i(A\times_C^hB)$ is the zero map (by exactness of \eqref{2}) and we get a SES
$$
0\to H^{i}\left(A \times_{C}^{h} B\right) \xrightarrow{H^i(p)\times H^i(q)} H^{i}(A) \oplus H^{i}(B) \xrightarrow{H^i(f)-H^i(g)} H^{i}(C)\to 0.
$$
This SES is split as $0\times (-H^i(g)^{-1})$ is a section of $H^i(f)-H^i(g)$.
And here is where I get stuck. How can I conclude that $H^i(p)$ is an isomorphism using that $H^i(g)$ is an iso? I was trying to prove it following the proof of the splitting lemma but that didn't work.
 A: Here is the step missing:

Lemma. Let $\mathcal{A}$ be an abelian category, and let $f:A\to B$, $g:A\to C$, $h:B\to D$, and $k:C\to D$ be morphisms in $\mathcal{A}$. It holds:

*

*If the sequence
$$
0\to A\xrightarrow{f\times g}B\oplus C\xrightarrow{h\amalg k}D\to 0
$$
is short exact, then $k$ being an isomorphism implies $f$ is an isomorphism.

*$\operatorname{Im}(\operatorname{Im}(f\times g)\to B\oplus C\xrightarrow{p} B)=\operatorname{Im}f$, where $p$ is the canonical projection.


Proof. For proving 1 we will need 2, so we prove 2 first. Note the result is obvious for $\mathcal{A}=\mathsf{Ab}$ the category of abelian groups, but in the general case we have to work with universal properties of (co)kernels.
Denote $h$ to the composite $\operatorname{Im}(f\times g)\to B\oplus C\xrightarrow{p} B$, so we want to show that $\operatorname{Im}h=\operatorname{Im}f$. We claim that the composite $\operatorname{Im}f\to B\to\operatorname{Coker}h$ vanishes. We can verify nullity after pre-composition with $A\to\operatorname{Im}f$, as the latter is epic (here we are using abelianity of $\mathcal{A}$). By commutativity of the following diagram, the claim follows from the fact that $(\operatorname{coker}h)\cdot h=0$.

Now we show $\operatorname{Im}f$ shows the universal property of an image of $h$. Suppose $D\to B$ is a map such that the composite $D\to B\to\operatorname{Coker} h$ vanishes. It suffices to show that there exists a dotted arrow $\operatorname{Coker}h\dashrightarrow\operatorname{Coker}f$ that makes the following diagram commutative

because on that case, the morphism $D\to B$ will factor through $\operatorname{Im}f$ since it is the kernel of $B\to\operatorname{Coker}f$. Uniqueness of the factorization will come from the fact that $\operatorname{Im}f\to B$ is monic (since it is a kernel). To construct the dotted arrow, we use the universal property of $\operatorname{Coker}h$. To verify that $(\operatorname{coker} f)pi$ vanishes, we can verify it after pre-composition with $k$ (again, by abelianity of $\mathcal{A}$, $k$ is a surjection), and we have $(\operatorname{coker} f)pik=(\operatorname{coker} f)p(f\times g)=(\operatorname{coker} f)f=0$.
We now prove 1. It suffices to prove the claim for $C=D$ and $k=-1_C$, as we have an isomorphism of short exact sequences
$\require{AMScd}$
$$
\begin{CD}
0@>>> A@>{f\times g}>> B\oplus C@>{h\amalg k}>>D@>>>0\\
@.@|@|@VV{-k^{-1}}V\\
0@>>> A@>>{f\times g}> B\oplus C@>>{(-hk^{-1})\amalg(-1_C)}>C@>>> 0
\end{CD}
$$
We show then that if
$$
\tag{3}\label{3}
0\to A\xrightarrow{f\times g}B\oplus C\xrightarrow{h\amalg(-1_C)}C\to 0
$$
is short exact, then $f$ is an isomorphism. With the purpose of understand why the result is true, we show it first for $\mathcal{A}=\mathsf{Ab}$ the category of abelian groups. Really the general proof is just the categorification of this particular case. If \eqref{3} is exact, we have that
$$
\tag{4}\label{4}
\{(f(a),g(a))\mid a\in A\}
=\operatorname{Im}(f\times g)
=\operatorname{Ker}(h\amalg(-1_C))
=\{(b,h(b))\mid b\in B\}.
$$
From the equality of the leftmost side with the rightmost one, it follows at once that $f$ is surjective. But from this equality it also follows that $g(a)=h(f(a))$ for all $a\in A$, so $g=hf$. From here we deduce that $f$ is injective, since $f\times g$ is injective: $f(a)=f(a')$ implies
$$
(f\times g)(a)=(f(a),g(a))=(f(a),h(f(a)))=(f(a'),h(f(a')))=(f\times g)(a')
$$
which in turn implies $a=a'$. Hence $f$ is bijective.
Now we give the proof for a general abelian category $\mathcal{A}$. We cannot consider elements anymore, but we can still work with universal properties. As we did before, we will show first that $f$ is onto and then that it is injective. From the fact that $\mathcal{A}$ is balanced (abelian categories are), it will follow that $f$ is an isomorphism.
We need an abelian-category-theoretic way of expressing the RHS of \eqref{4}. This RHS corresponds to $\operatorname{Im}(1_B\times h)$, so we will show that
$$
\tag{5}\label{5}
\operatorname{Ker}(h\amalg(-1_C))=\operatorname{Im}(1_B\times h).
$$
We show that $\operatorname{Im}(1_B\times h)$ is a kernel of $h\amalg(-1_C)$. For that, we first have to show that the composite $\operatorname{Im}(1_B\times h)\to B\oplus C\xrightarrow{h\amalg(-1_C)}C$ vanishes. But this is easy, using the image factorization of $1_B\times h$ and the fact that $B\to\operatorname{Im}(1_B\times h)$ is surjective as $\mathcal{A}$ is abelian, as $(h\amalg(-1_C))\cdot(1_B\times h)=h-h=0$.

We prove that the RHS of \eqref{5} satisfies the UP of the LHS. Suppose $k\times\ell:D\to B\oplus C$ is a morphism such that $0=(h\amalg(-1_C))\cdot(k\times\ell)=hk-\ell$. Thus $\ell=hk$.
We get a solid commutative diagram

By commutativity of the right upper triangle, the vertical composite $D\xrightarrow{k\times\ell}B\oplus C\to \operatorname{Coker}(1_B\times h)$ vanishes, so $k\times\ell$ factors through $\operatorname{Im}(1_B\times h)$ via the dotted arrow of the last diagram. Uniqueness of the factorization comes from the fact that $\operatorname{Im}(1_B\times h)\to B\oplus C$ is monic.
This shows \eqref{5}. Using this equation and exactness of \eqref{3}, we deduce that
$$
\operatorname{Im}(f\times g)=\operatorname{Im}(1_B\times h).
$$
From this last equality plus point 1 of the lemma, it follows that $f$ and $1_B$ have same image, i.e., $\operatorname{Im}f=B$ and $f$ is surjective (this is the substitute of the inference we did in the case of abelian groups to conclude surjectiveness then). It is left to see that $f$ is injective. As it was the case before, we claim that $hf=g$ holds now as well. To see this, note that $1_B\times h$ is injective (in abelian categories, the product of a monomorphism with any other map is a monomorphism), so $\operatorname{im}(1_B\times h)=1_B\times h$, as in a preabelian category the image of a monomorphism is the map itself. Hence, we can use the image factorization of $f\times g$ to get a factorization

That is, $f\times g=(1_B\times h) f'=f'\times hf'$. Therefore $f=f'$ and $g=hf$.
Lastly, since $f\times g=f\times hf=(1_B\times h)f$ is injective, we deduce that $f$ is also injective. $\square$
A: Okay, I've found an easier proof of the same lemma from my last answer, which I here restate

Lemma. Let $\mathcal{A}$ be an abelian category, and let $f:A\to B$, $g:A\to C$, $h:B\to D$, and $k:C\to D$ be morphisms in $\mathcal{A}$, and suppose the sequence
$$
0\to A\xrightarrow{f\times g}B\oplus C\xrightarrow{h\amalg k}D\to 0
$$
is short exact. Then $k$ being an isomorphism implies $f$ is an isomorphism.

Proof. As we argued in the other answer, without loss of generality we may assume $k=1_C$, since we have an isomorphism of short exact sequences
we have an isomorphism of sequences
$\require{AMScd}$
$$
\begin{CD}
0@>>> A@>{f\times g}>> B\oplus C@>{h\amalg k}>>D@>>>0\\
@.@|@|@VV{k^{-1}}V\\
0@>>> A@>>{f\times g}> B\oplus C@>>{(hk^{-1})\amalg(1_C)}>C@>>> 0
\end{CD}
$$
But we also have an isomorphism of complexes
$\require{AMScd}$
$$
\tag{$\star$}\label{diag}
\begin{CD}
0@>>> A@>{\begin{pmatrix} f\\ g\end{pmatrix}}>> B\oplus C@>{(h,1_C)}>>C@>>>0\\
@.@|@VV{\begin{pmatrix} 1_B&0\\ h&1_C\end{pmatrix}}V@|\\
0@>>> A@>>{\begin{pmatrix} f\\ 0\end{pmatrix}}> B\oplus C@>>{(0,1_C)}>C@>>> 0
\end{CD}
$$
Here we are using matrix calculus. The matrix in the middle is invertible for its inverse is $\begin{pmatrix} 1_B&0\\ -h&1_C\end{pmatrix}$.
In \eqref{diag} the three vertical morphisms indeed define a chain map, since the right square clearly commutes and the left square commutes as $0=hf+g$, for the top sequence is a complex.
Therefore, in \eqref{diag}, if the top complex is short exact so is the bottom one. On this case, the kernel of $(0,1_C)$, i.e., $\begin{pmatrix} 1_B\\ 0\end{pmatrix}:B\to B\oplus C$, by exactness coincides with the image of $\begin{pmatrix} f\\ 0\end{pmatrix}$. Therefore

is the factorization of $\begin{pmatrix}f\\0\end{pmatrix}$ through its image. Since $\begin{pmatrix}f\\0\end{pmatrix}$ is injective and $\mathcal{A}$ is abelian, the morphism into the image, $f$, becomes an isomorphism.
