Let $C$ be an abelian category with enough injectives and $F$ be a left exact additive functor. Consider the short exact sequence $0 \to A' \to A \to A ''\to 0$. Therefore, we have connecting maps $\delta_i: R^i F(A'') \to R^{i+1}F(A')$.

Now let $0 \to A' \to I^\bullet$ be an injective resolution of $A'$. The resolution $0 \to A' \to A \to A''\to 0$ maps to $0 \to A' \to I^\bullet$ (unique upto homotopy). So we get a map $\phi:A'' \to I^1$ such that $A''$ lands inside $\text{ker} (I^1 \to I^2)$. So this gives a map from $F(A'') \to R^1 F(A')$, which nfdc23 mentioned here to be negative the first connecting map.

Now let us choose an injective resolution $0 \to A'' \to K^\bullet$. The map $\phi:A'' \to I^1$ can be extended to obtain a map $K^0 \to I^1$. This extends and gives maps $K^n \to I^{n+1}$ which gives a map of complexes $K^\bullet \to I^{\bullet {+1}}$. Applying $F$ and taking cohomology gives maps $\phi_i:R^i F (A'') \to R^{i+1}F(A')$.

Question: What is the relation between $\delta_i $ and $\phi_i$? Does it follow that $\delta_i = (-1)^{i+1} \phi_i$?

An approach: Let us choose injective resolutions $0 \to A' \to I^\bullet$, $0 \to A \to J^\bullet$, $0 \to A'' \to K^\bullet$ which fit into an exact sequence of complexes. Now the exact sequences $0 \to I^n \to J^n \to K^n \to 0$ splits. Let us denote the splitting maps by $r_n: J^n \to I^n$ and $s_n: K^n \to J^n$.

Now the "connecting maps" correspond to the composition: $\text{ker}(K^n \to K^{n+1}) \to J^n \to J^{n+1} \to I^{n+1}$, where the first map is given by $s_n$ and last map is by $r_{n+1}$. Therefore, the answer seem to depend on the possibility of choosing spliting maps with certain commutative/anticommutative properties.

  • $\begingroup$ I have a proof of this now, which answers this positively. $\endgroup$ Dec 7, 2016 at 19:36


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