Quasi isomorphism and pullback

Question

Let $A,B,C$ be chain complex, $f : A \to B$ a quasi isomorphism and $g : C \to B$ a cochain map. Consider the pullback $A \times_B C$ with its two projections $p_A$ and $p_C$. Is it possible to show that $p_C$ is a quasi isomorphism without diagram chase ?

Thanks a lot !!

Answer 1

The typical result is that in a model category, weak equivalences are preserved by pullbacks along fibrations, not any random maps (though sometimes the class of maps along which weak equivalences are preserved is bigger, see e.g. this blog post by Mike Shulman).

Let's say we are considering nonnegatively graded chain complexes. Then a standard result is that there's a model structure on $\mathsf{Ch}_{\ge 0}(R)$ (for some ring $R$) in which weak equivalences are quasi-isomorphisms and fibrations are chain maps that are surjective in positive degrees (cf. standard textbooks, for examples Hovey's book Model categories).

The upshot, in your situation, is that if you require $g$ to be surjective in positive degrees, then $p_C$ will be a quasi-isomorphism again. You can prove it using the machinery of model categories, which can qualify as a "proof without diagram chasing" (though note that proving $\mathsf{Ch}_{\ge 0}(R)$ is a model category involves diagram chasing, so I'm not sure what you've won this way). You can also give a down-to-earth proof here, but it does involve diagram chasing, sorry. But as you'll see it's not so bad...

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To fix ideas, consider the diagram: $$\require{AMScd} \begin{CD} A \times_B C @>p>> C \\ @VVV @VgVV \\ A @>f>\sim> B \end{CD}$$ and assume that $g$ is surjective in positive degrees and that $f$ is a quasi-isomorphism. Let's prove that $p$ is a quasi-isomorphism.

Injectivity

Let $(x,y) \in A_n \times_{B_n} C_n$ be a cycle ($dx = 0$, $dy = 0$) such that $p(x,y) = y = dz$ for some $z \in C_{n+1}$. In other words, $[x,y] \in H_n(A \times_B C)$ is a homology class such that $p_*[x,y] = 0$. We want to show that $(x,y)$ is a boundary too (so $[x,y] = 0$).
Since $f(x) = g(y) = g(dz) = d(g(z))$ is a boundary and $f$ is a quasi-isomorphism, $x$ is a boundary too: $x = du$ for some $u \in A_{n+1}$. Thus $d(u,z) = (x,y)$ is a boundary. Conclusion: $p_*$ is injective on $H_n$.

Surjectivity

Let $y \in C_n$ be some cycle ($dy=0$). We are looking for a homology class $[w] \in H_n(A \times_B C)$ such that $p_*[w] = [y]$, i.e. $p(w) = y + d(\dots)$.
Since $y$ is a cycle, $g(y) \in B_n$ is also a cycle, and $f$ is a quasi-iso, so there exists a cycle $x \in A_n$ and some $z \in B_{n+1}$ such that $g(y) = f(x) + dz$ (i.e. $f_*[x] = [g(y)]$). Since $g$ is surjective in positive degrees (and $n+1 > 0$), there's some $u \in C_{n+1}$ such that $g(u) = z$, hence: $$g(y) = f(x) +dz = f(x) + g(du) \implies g(y - du) = f(x).$$ So consider $w = (x, y-du) \in A_n \times_{B_n} C_n$. It's a cycle, and $p(w) = y + du$ is homologous to $y$. Conclusion: $p$ is surjective on $H_n$.

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As you can see, I really used the hypothesis that $g$ was surjective in positive degrees. In general, without that assumption, $p_*$ will not necessarily be surjective. If you are working with unbounded chain complexes, a fibration is a surjective (in all degrees) chain map, and the proof is essentially the same. (Cofibrations are much harder to describe too.)