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Let $\mathcal{A}$ be an abelian category (for simplicity you can think that $\mathcal{A}$ is the category of modules over ring $R$).

Let $[1]$ be the category with two objects and one arrow between them.

$$\bullet \rightarrow \bullet $$

Denote by $\hat{\mathcal{A}}$ the category of functors from $[1]$ to $\mathcal{A}$.

$\hat{\mathcal{A}}$ is an abelian category as well.

There is a functor $ \ker: \hat{\mathcal{A}} \rightarrow \mathcal{A}$ defined as follows. An object of $\hat{\mathcal{A}}$ is pair of objects in $\mathcal{A}$ and a map between them $$\hat{A} = (A_1 , A_2, f: A_1 \rightarrow A_2).$$

The functor $\ker$ is defined by the formula $$ \ker ( \hat{A} ) = \ker(f).$$

It is easy to see that $\ker$ is left-exact. There are enough injectives in $\hat{\mathcal{A}}$. So we can define $R^i \ker$.

Question: What are $R^i \ker$ explicitly?

Comment 1. I expect the answer to be given in terms of classical constructions in the category $\mathcal{A}$ (like the $\operatorname{Ext}$ functor).

Comment 2 I found the answer in case of $A = \mathrm{Vect}_{\mathbb{k}}$: \begin{align*} R^1 \ker &= \text{coker}, \\ R^i \ker &= \text{$0$ for $i > 1$}. \end{align*}

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It's exactly what you said in any abelian category: $R^1$ is the cokernel and the higher derived functors vanish. This is more or less the content of the snake lemma.

One reason to believe this is that taking $\text{ker}$ of a map is the same thing as taking the zeroth cohomology of a two-term cochain complex.

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  • $\begingroup$ You constructed delta-functor. To complete proof you have to prove that it is universal delta-functor. $\endgroup$ – Peter Cotowsky May 8 '15 at 20:12

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