Taking cohomology commutes with any exact functor

In a previous positing, we have talked about why taking Čech cohomology of a quasi-compact separated scheme $X$ over a field $k$ commutes with base change along any field extension $k \hookrightarrow K.$ If one carefully reads the proof,  it is not difficult to note that we have proved that given any cochain $$C : 0 \rightarrow C^{0} \rightarrow C^{1} \rightarrow C^{2} \rightarrow \cdots$$ of $k$-vector spaces, if $$C \otimes_{k} K : 0 \rightarrow C^{0} \otimes_{k} K \rightarrow C^{1} \otimes_{k} K \rightarrow C^{2} \otimes_{k} K \rightarrow \cdots$$ denotes the induced cochain of $K$-vector spaces by the tensor product, then we have $$H^{i}(C \otimes_{k} K) \simeq H^{i}(C) \otimes_{k} K$$ for all $i \geq 0.$ There are more things to keep in mind from the proof (in the previous positing):

• we don't have to assume that $C^{i} = 0$ for $i < 0$;
• the fact that $H^{i}(C) \otimes_{k} K = (\ker(d^{i}) \otimes_{k} K)/\mathrm{im}(d^{i-1} \otimes_{k} K)$ only used the fact that $(-) \otimes_{k} K$ is a right-exact functor;
• exactness of $(-) \otimes_{k} K$ (as $k$ is a field) is used to ensure that the canonical map $\ker(d^{i}) \otimes_{k} K \rightarrow \ker(d^{i} \otimes_{k} K)$ is an isomorphism, which resulted in $$\begin{align*}H^{i}(C) \otimes_{k} K &= (\ker(d^{i}) \otimes_{k} K)/\mathrm{im}(d^{i-1} \otimes_{k} K) \\ &\simeq \ker(d^{i} \otimes_{k} K)/\mathrm{im}(d^{i-1} \otimes_{k} K) \\ &= H^{i}(C \otimes_{k} K).\end{align*}$$

In particular, our observations immediately show that if $C$ is any cochain complex of $A$-modules over a (commutative unital) ring $A$ and $B$ is any $A$-algebra, then for each $i \in \mathbb{Z},$ we always have a map $$H^{i}(C) \otimes_{A} B \rightarrow H^{i}(C \otimes_{A} B),$$ and this map is an ismorphism if $B$ is flat over $A$ (i.e., if the functor $(-) \otimes_{A} B : \textbf{Mod}_{A} \rightarrow \textbf{Mod}_{B}$ is exact). Let's prove a more general statement.

Theorem. Given two rings $A, B,$ let $F : \textbf{Mod}_{A} \rightarrow \textbf{Mod}_{B}$ be any additive functor and fix any cochain complex $$C : \cdots \overset{d^{i-2}}{\longrightarrow} C^{i-1} \overset{d^{i-1}}{\longrightarrow} C^{i} \overset{d^{i}}{\longrightarrow} C^{i+1} \overset{d^{i+1}}{\longrightarrow} \cdots$$ in the category $\textbf{Mod}_{A}$ of $A$-modules. Denote by $F(C)$ the cochain complex of $B$-modules given by applying the additive functor $F$ to the cochain compelx $C$ of $A$-modules.

1. If $F$ is right-exact, then we have a canonical $B$-linear map $F(H^{i}(C)) \rightarrow H^{i}F(C)$ for each $i \in \mathbb{Z}.$
2. If $F$ is left-exact, then we have a canonical $B$-linear map $F(H^{i}(C)) \leftarrow H^{i}F(C)$ for each $i \in \mathbb{Z}.$
3. If $F$ is exact, then for each $i \in \mathbb{Z},$ the maps given above are inverses to each other so that $$F(H^{i}(C)) \simeq H^{i}(F(C)).$$

Remark. Note that we did not even mention what a "canonical map" means. In this case, one needs to look at the proof to see how such a map is induced and figure out in what sense it is canonical.

Before proving the theorem, let us not assume left-exactness nor right-exactness for $F.$

1st map. We may apply $F$ to the exact sequence $$C^{i-1} \overset{d^{i-1}}{\longrightarrow} \ker(d^{i}) \overset{\pi_{i}}{\longrightarrow} \ker(d^{i})/\mathrm{im}(d^{i-1}) = H^{i}(C) \rightarrow 0$$ to get the following cochain: $$F(C^{i-1}) \overset{F(d^{i-1})}{\longrightarrow} F(\ker(d^{i})) \overset{F(\pi_{i})}{\longrightarrow} F(\ker(d^{i})/\mathrm{im}(d^{i-1})) = F(H^{i}(C)).$$ By the universal property of the cokernel map $$F(\ker(d^{i}))\overset{\rho_{i}}{\longrightarrow} \frac{F(\ker(d^{i}))}{\mathrm{im}(F(d^{i-1}))} \rightarrow 0,$$ there is a unique map $$\frac{F(\ker(d^{i}))}{\mathrm{im}(F(d^{i-1}))} \overset{\psi_{i}}{\longrightarrow} F(H^{i}(C))$$ such that $$F(\pi_{i}) = \psi_{i} \circ \rho_{i}.$$ In this sense, the map $\psi_{i}$ is canonical.

Remark. If $F$ is right-exact, the map $\psi_{i}$ is an isomorphism.

2nd map. We are still not assuming any exactness of $F.$ The exact sequence $$0 \rightarrow \ker(d^{i}) \overset{\iota_{i}}{\longrightarrow} C^{i} \overset{d^{i}}{\longrightarrow} C^{i+1}$$ gives rise to the following cochain: $$F(\ker(d^{i})) \overset{F(\iota_{i})}{\longrightarrow} F(C^{i}) \overset{F(d^{i})}{\longrightarrow} F(C^{i+1}).$$ This induces the cochian $$\frac{F(\ker(d^{i}))}{\mathrm{im}(F(d^{i-1}))} \overset{\overline{F(\iota_{i})}}{\longrightarrow} \frac{F(C^{i})}{{\mathrm{im}(F(d^{i-1}))}} \overset{\overline{F(d^{i})}}{\longrightarrow} F(C^{i+1}).$$ We note that $$0 \rightarrow H^{i}(F(C)) = \frac{\ker(F(d^{i}))}{{\mathrm{im}(F(d^{i-1}))}} \overset{\omega_{i}}{\longrightarrow} \frac{F(C^{i})}{{\mathrm{im}(F(d^{i-1}))}} \overset{\overline{F(d^{i})}}{\longrightarrow} F(C^{i+1})$$ is exact, so by the universal property of the kernel, there exists a unique map $$\frac{F(\ker(d^{i}))}{\mathrm{im}(F(d^{i-1}))} \overset{\phi_{i}}{\longrightarrow} H^{i}(F(C))$$ such that $$\overline{F(\iota_{i})} = \omega_{i} \circ \phi_{i}.$$ This uniqueness is why we say the map $\phi_{i}$ is canonical.

Remark. If $F$ is left-exact, the map $\phi_{i}$ is an isomorphism.

Proof. For the first statement, assume that $F$ is right-exact. Then the map $\psi_{i}$ is an isomorphism, so we get $$F(H^{i}(C)) \overset{\psi_{i}^{-1}}{\longrightarrow} \frac{F(\ker(d^{i}))}{\mathrm{im}(F(d^{i-1}))} \overset{\phi_{i}}{\longrightarrow} H^{i}(F(C)),$$ as desired.

For the second statement, assume that $F$ is left-exact. Then $\phi_{i}$ is left-exact, so we get $$H^{i}(F(C)) \overset{\phi_{i}^{-1}}{\longrightarrow} \frac{F(\ker(d^{i}))}{\mathrm{im}(F(d^{i-1}))} \overset{\psi_{i}}{\longrightarrow} F(H^{i}(C)),$$ as desired.

The third statement follows because if $F$ is exact, then both $\psi_{i}$ and $\phi_{i}$ are invertible and $$(\psi_{i}^{-1} \circ \phi_{i})^{-1} = \phi_{i}^{-1} \circ \psi_{i},$$ as desired. $\Box$

Non-mathematical remarks. Due to the formula $$F(H^{i}(C)) \simeq H^{i}(F(C)),$$ Ravi Vakil called this "FHHF Theorem" (Vakil, 1.6.H), and I believe in an earlier draft of his book, he called it "From Here Hop Far," which perhaps did not catch on. He also calls this Fernbahnhof(FernbaHnHoF), which seems like a German word, but I did not get it even after looking up what it meant.

More general situations. In fact, Vakil's book (1.6.H) formulates this in a much more general situation where an additive functor given between any two abelian categories.

Theorem. Given two abelian categories $\mathscr{A}, \mathscr{B},$ let $F : \mathscr{A} \rightarrow \mathscr{B}$ be any additive functor and fix any cochain complex $$C : \cdots \overset{d^{i-2}}{\longrightarrow} C^{i-1} \overset{d^{i-1}}{\longrightarrow} C^{i} \overset{d^{i}}{\longrightarrow} C^{i+1} \overset{d^{i+1}}{\longrightarrow} \cdots$$ in $\mathscr{A}$. Denote by $F(C)$ in $\mathscr{B}$ given by applying the additive functor $F$ to the cochain compelx $C$ in $\mathscr{A}.$

1. If $F$ is right-exact, then we have a canonical morphism $F(H^{i}(C)) \rightarrow H^{i}F(C)$ in $\mathscr{B}$ for each $i \in \mathbb{Z}.$
2. If $F$ is left-exact, then we have a canonical morphism $F(H^{i}(C)) \leftarrow H^{i}F(C)$ in $\mathscr{B}$ for each $i \in \mathbb{Z}.$
3. If $F$ is exact, then for each $i \in \mathbb{Z},$ the maps given above are inverses to each other so that $$F(H^{i}(C)) \simeq H^{i}(F(C)).$$

Proof. When we gave the proof for the case $\mathscr{A} = \textbf{Mod}_{A}$ and $\mathscr{B} = \textbf{Mod}_{B},$ all the maps are induced by certain universal properties, so this should work in general as well. $\Box$