Let $\mathcal{A}$, $\mathcal{B}$ be abelian categories, and let $F^\bullet: \mathcal{A} \rightarrow \mathcal{B}$ be a cohomological $\delta$-functor.
Recall that a cohomological $\delta$-functor is a series of additive functors $F^{n}: \mathcal{A} \rightarrow \mathcal{B}$ for $n \geq 0$ along with morphisms $\delta_{ABC}^n: F^n{C} \rightarrow F^{n+1}{A}$ for any short exact sequence $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$ in $\mathcal{A}$. A cohomological $\delta$-functor must satisfy these two properties
- For any short exact sequence $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$ in $\mathcal{A}$, there is a long exact sequence
$$ 0 \rightarrow F^0(A) \rightarrow F^0(B) \rightarrow F^0(C) \xrightarrow{\delta_{ABC}^0} F^1(A) \rightarrow F^1(B)... $$
- If we have a morphism of long exact sequences
$$ \require{AMScd} \begin{CD} 0 @>>> A @>>> B @>>> C @>>> 0 \\ @. @VVV @VVV @VVV @. \\ 0 @>>> A' @>>> B' @>>> C' @>>> 0 \end{CD} $$
then the following diagram commutes.
$$ \require{AMDcd} \begin{CD} F^n(C) @>{\delta^n_{ABC}}>> F^{n+1}(A) \\ @VVV @VVV \\ F^n(C') @>{\delta^n_{A'B'C'}}>> F^{n+1}(A') \\ \end{CD} $$
Suppose that we have a commuting square of short exact sequences:
$$ \require{AMScd} \begin{CD} @. 0 @. 0 @. 0 @. \\\ @VVV @VVV @VVV @. \\ 0 @>>> A @>>> B @>>> C @>>> 0 \\ @. @VVV @VVV @VVV @. \\ 0 @>>> D @>>> E @>>> F @>>> 0 \\ @. @VVV @VVV @VVV @. \\ 0 @>>> H @>>> I @>>> J @>>> 0 \\ @. @VVV @VVV @VVV @. \\ @. 0 @. 0 @. 0 @. \end{CD} $$
Then, does this square anti-commute?
$$ \require{AMDcd} \begin{CD} F^n(J) @>{\delta^n_{HIJ}}>> F^{n+1}(H) \\ @VV{\delta^n_{CFJ}}V @VV{\delta^{n+1}_{ADH}}V \\ F^{n+1}(C) @>{\delta^{n+1}_{ABC}}>> F^{n+2}(A) \\ \end{CD} $$
Cohomological $\delta$-functors are meant to generalize other "cohomology" functors found in homological algebra, such as the cohomology of a cochain complex and right-derived functors. In both of these examples, the square above does anticommute. But I have not been able to solve prove it for general cohomological $\delta$-functors.
