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Tuesday, August 13, 2019

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

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