Long exact sequence in cohomology associated to a short exact sequence of *functors*

In homological algebra, when you have a left exact functor $F$ From an abelian category $\mathcal{A}$ to an abelian category $\mathcal{B}$ and you have enough injectives in $\mathcal{A}$, then you can define the right derived functors of $F$ by applying $F$ to an injective resolution of an object $A$ and calculating the cohomologies of the resulting chain complex. I will use the non-standard notation $H^n(F,A)$ for this.

It is a fundamental property of the derived functors that given a short exact sequence of objects $$ 0 \to A\to B\to C\to 0 $$

in $\mathcal{A}$, you get a long exact sequence in cohomology $$ ...\to H^n(F,A)\to H^n(F,B)\to H^n(F,C)\to H^{n+1}(F,A)\to... $$

Now, it seems to me that there is a dual thing going on for a short exact sequence of functors. Namely, If you have a short exact sequence of Left exact functors $$ 0\to F\to T\to S\to 0 $$

i.e. natural transformations such that for each hobject $X$ we obtain a short exact sequence, then for each object $X$ we obtain a long exact sequence in cohomology

$$ ...\to H^n(F,X)\to H^n(T,X)\to H^n(S,X)\to H^{n+1}(F,X)\to... $$

The point is that applying the short exact sequence of functors to an injective resolution of $X$ gives a short exact sequence of chain complexes hence a long exact sequence in cohomology. Moreover, It seems to me that if $F,T$ are left exact then a diagram chase shows that $S$ is also left exact. My question is

Is this all correct? If it is (or approximately at least) what is the "correct" way to present this duality and where can I read about it?

As a concrete example consider a group $G$ and a subgroup $K$, we can define the functors of $G$ and $K$ fixed points from the category of $G$-modules to the category of abelian groups. Both are left exact and if we set $S(A) = A^K/A^G$ we obtain a short exact sequence of left exact functors. This shows that the restriction homomorphisms $H^n(G,A)\to H^n(K,A)$ fit into a long exact sequence $$ ...\to H^(G,A)\to H^n(K,A)\to H^n(S,A)\to H^{n+1}(G,A)\to... $$

The groups $H^n(S,A)$ could (should?) be called the "relative group cohmology" and perhaps denoted something like $H^n(G,K,A)$ (like in topology).

I did not see this anywhere (but also didn't look too hard). I would be happy to hear a systematic explanation of my mistakes...

Edit: One mistake is my claim that if $F,T$ are left exact then so is $S$. This is not true. I guess there should be a "sheafification" operations that left-exactify the quotient. In other words, we should take a short exact sequence in the category of left exact functors.

Solution 1:

Yes, right derived functors $R^nTA$ come automatically with two long exact sequences, one from varying $A$ and one from varying the functor $T$. This is discussed in chapter IV.6. of Hilton and Stammbach's book A Course in Homological Algebra. What follows is essentially me summarizing the content of this chapter.

Your notion of exactness of a sequence of functors is not the most efficient one. Let's say a sequence of natural transformations $T'\to T\to T''$ is short exact at injectives if and only if \begin{align*}0\to T'I \to TI\to T''I\to 0\end{align*} is exact for every injective object $I$. Given a sequence of functors as above which is short exact at injectives, fix an injective resolution $A\to \mathbf I$ of $A$. The natural transformations yield a sequence of chain maps \begin{align*} 0\to T'\mathbf I \to T\mathbf I \to T''\mathbf I \to 0\end{align*} which is by assumption levelwise short exact and thus short exact. Hence we get a long exact sequence in cohomology. \begin{align*} 0 \to R^0T'A \to R^0TA \to R^0T''A\xrightarrow{\omega_0} R^1T'A \to R^1TA\to R^1T''A\to...\end{align*} The long exact sequence is natural in $T$ and in $A$ in the sense desecribed in Hilton & Stammbach IV.6. Proposition 6.4. You do not need to know that the functors $T',T$ and $T''$ are left exact to get the long exact sequence. Your example does work. The most important examples are actually the classical derived functors $\mathrm{Tor}$ and $\mathrm {Ext}$.

Example. Let $\mathcal A$ be any abelian category with enough injectives. Define $\mathrm{Ext}_{\mathcal A}^n(A,-) $ as the right derived functor of $\mathrm{Hom}_{\mathcal A}(A,-)$. In other words \begin{align*} \mathrm{Ext}_{\mathcal A}^n(A,B) = R^n\mathrm{Hom}_{\mathcal A}(A,-)B\,.\end{align*} Given a short exact sequence $0\to B'\to B\to B''\to 0$ you obtain a long exact sequence in the second variable by the standard argument involving the Horseshoe lemma. But we can obtain also a long exact sequence in the second variable by using the method described above. A morphism $A'\to A$ yields a natural transformation $\mathrm{Hom}_{\mathcal A}(A',-)\to \mathrm{Hom}_{\mathcal A}(A,-)$. Thus a short exact sequence $0\to A'\to A\to A''\to 0$ yieds a sequence of natural transformations \begin{align*} \mathrm{Hom}_{\mathcal A}(A'',-) \to \mathrm{Hom}_{\mathcal A}(A,-)\to \mathrm{Hom}_{\mathcal A}(A',-) \end{align*} This sequence is short exact at injectives by definition of injective objects (but it is not short exact in your sense!). Applying the discussion above to the sequence yields the long exact sequence of $\mathrm{Ext}_{\mathcal A}^n$ in the first variable. In a similar way one can show that $\mathrm{Tor}_n^\Lambda(A,B) = L_n(A\otimes_\Lambda -)B$ has two kinds of long exact sequences. One is obtained by varying $A$ and one by varying $B$.

Additionally, since a morphism $A\to A'$ induces a natural transformation $\textrm{Hom}_{\mathcal A}(A',-) \to \textrm{Hom}_{\mathcal A}(A,-)$ and thus a map $\mathrm{Ext}_{\mathcal A}^n(A',B) \to \mathrm{Ext}_{\mathcal A}^n(A,B)$ we have found an easy way to show that $\mathrm{Ext}_{\mathcal A}^n$ is a bifunctor, without having to derive $\mathrm{Hom}_{\mathcal A}$ in the first variable and having to show that $\mathrm{Ext}_{\mathcal A}^n$ is balanced first. I have seen people doing it this way, and it is a mess. Both of the two long exact sequences of $\mathrm{Ext}_{\mathcal A}^n$ vary naturally in both $A$ and $B$.