# Example of an additive functor admitting no right derived functor, 2

This is a sequel to the question

the purpose being to state a particular case of  which would be as easy as possible (though this particular case is still much too hard for me).

See  for any unexplained notation and for some of the definitions involved.

Let $k$ be a field, let $T$ be an indeterminate, let $A$ be the $k$-algebra $k[[T]]$, let $\mathcal C$ be the category opposite to the category of $A$-modules, let $\mathcal C_0$ be the full subcategory of finite dimensional objects of $\mathcal C$, and let $\mathcal C'$ be the category opposite to the category of $k$-vector spaces. Denote again by $k$ the residue field of $A$, so that $k\in\mathcal C_0$. Let $F:\mathcal C\to\mathcal C'$ be the functor $\text{Hom}_A(k,\ )$, let $F_0:\mathcal C_0\to\mathcal C'$ be its restriction, let $RF:\text D(\mathcal C)\to\text D(\mathcal C')$ be the right derived functor of $F$ (it exists by Theorem 14.4.3 p. 359 in the book Categories and Sheaves by Kashiwara and Schapira), and let $(RF)_0:\text D(\mathcal C_0)\to\text D(\mathcal C')$ be its restriction.

Assume that $RF_0:\text D(\mathcal C_0)\to\text D(\mathcal C')$ exists. My hope is to derive a contradiction from this assumption, but I'm still far from this. A weaker goal would be to prove that the natural morphism $RF_0\to(RF)_0$ is not an isomorphism.

Here is a first (very modest) step in this direction. For $X$ in $\text K(\mathcal C)$ let $\text{Qis}^X$ be the category of quasi-isomorphisms $X\to Y$ with $Y$ in $\text K(\mathcal C)$, and, for $X$ in $\text K(\mathcal C_0)$ let $\text{Qis}_0^X$ be the category of quasi-isomorphisms $X\to Y$ with $Y$ in $\text K(\mathcal C_0)$. If $X$ is in $\text K(\mathcal C_0)$, then $\text{Qis}_0^X$ is not necessary cofinal to $\text{Qis}^X$.

To prove this let $X\in\text K(\mathcal C_0)$ be equal to $k$ in degree 0 and to 0 in the other degrees, let $Y\in\text K(\mathcal C)$ be equal to $A$ in degree 0, to the maximal ideal $\mathfrak m$ of $A$ in degree 1, and to 0 in the other degrees, the differential $A\to\mathfrak m$ being the obvious epimorphism, and let $X\to Y$ be the obvious quasi-isomorphism. It is easy to see that there is no quasi-isomorphism $Y\to Z$ in $\text K(\mathcal C)$ with $Z$ in $\text K(\mathcal C_0)$. This shows that $\text{Qis}_0^X$ is not cofinal to $\text{Qis}^X$.

Again, my question is:

Assuming that $RF_0:\text D(\mathcal C_0)\to\text D(\mathcal C')$ exists, is the natural morphism $RF_0\to(RF)_0$ an isomorphism?

(It would probably better to consider an additive functor $F$ of abelian categories such that there is no obvious candidate for $RF$, but I haven't found such an example.)

• Here is another additive functor $F$ of abelian categories for which one can hope to prove that it admits no right derived functor: Let $k$ and $A$ be as above, and let $\text{Mod}_{\text{fd}}(A)$ and $\text{Mod}_{\text{fd}}(k)$ be the categories of finite dimensional $A$-modules and $k$-vector spaces respectively. Then $F:\text{Mod}_{\text{fd}}(A)\to\text{Mod}_{\text{fd}}(k)$ is the additive functor $\text{Hom}_A(k,\ )$. – Pierre-Yves Gaillard Mar 18 '17 at 23:33