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I asked the same question a week ago on Mathematics Stackexchange but got no answer.

What would be a simple example of an additive functor $F:\mathcal C\to\mathcal C'$ of abelian categories such that the right derived functor $$ RF:\text D(\mathcal C)\to\text D(\mathcal C') $$ does not exist?

My reference for the notions involved in this post is the book Categories and Sheaves by Kashiwara and Schapira.

Here is a reminder of the definition of a right derived functor $RF$ in the above setting:

Let $\text K(F):\text K(\mathcal C)\to\text K(\mathcal C')$ be the triangulated functor induced by $F$ between the homotopy categories, let $X$ be in $\text K(\mathcal C)$, assume that the colimit $$ \operatorname*{colim}_{X\to Y}\ \text K(F)(Y),\tag1 $$ where $X\to Y$ runs over all the quasi-isomorphisms out of $X$ in $\text K(\mathcal C)$, exists in $\text D(\mathcal C')$, and denote this colimit by $RF(X)$. The right derived functor $RF$ of $F$ is defined at $X\in\text D(\mathcal C)$ if, for any functor $G:\text D(\mathcal C')\to\mathcal A$, the colimit $$ \operatorname*{colim}_{X\to Y}\ G(\text K(F)(Y)) $$ exists in $\mathcal A$ and coincides with $G(RF(X))$. The right derived functor $RF$ of $F$ exists if $RF$ is defined at $X$ for all $X$ in $\text D(\mathcal C)$.

The ideal would be to have an example of a pair $(F,X)$, where $F:\mathcal C\to\mathcal C'$ is an additive functor of abelian categories and $X$ is an object of $\text D(\mathcal C)$, such that (1) does not exist in $\text D(\mathcal C')$.

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Let ${\cal C}$ be the category of finite dimensional ${\bf Z}/2$-vector spaces equipped with a ${\bf Z}/2$ action, let ${\cal C'}$ be the category of finite dimensional ${\bf Z}/2$-vector spaces and let $F: {\cal C} \to {\cal C'}$ be the fixed subspace functor $V \mapsto V^{{\bf Z}/2}$. Consider the chain complex $X$ such that $X_n = {\bf Z}/2$ with trivial action in each degree $n \in {\bf Z}$ and with trivial differentials. Then the colimit $RF(X) = {\rm colim}_{X \to Y}Y^{{\bf Z}/2}$ does exist as a complex of vector spaces (since this category has all limits and colimits), and $RF(X)$ will be the derived functor there, but it will not be degreewise finite dimensional (and from here a simple argument shows that the colimit cannot exist in ${\cal D}({\cal C'})$). In fact, since ${\rm Ext^n_{{\cal C}}({\bf Z}/2,{\bf Z}/2)} \cong {\bf Z}/2$ for all $n \geq 0$ a spectral sequence argument shows that $RF(X)$ has infinite dimensional homologies in all degrees.

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  • $\begingroup$ It seems to me we can argue as follows. Assume by contradiction that $RF$ exists. The proof of Theorem 13.4.1 p. 337 in Kashiwara and Schapira's book shows $$R_0F(X)\cong\prod{\rm Hom}_{{\cal D}({\cal C})}({\bf Z}/2[0],{\bf Z}/2[n]),$$ which is infinite dimensional over ${\bf Z}/2$, a contradiction. What do you think? $\endgroup$
    – Pierre-Yves Gaillard
    Commented Mar 21, 2017 at 18:53
  • $\begingroup$ [When you say "$RF(X)$ will be the derived functor there" you don't take into account, it seems to me, the condition on the arbitrary functor $G$ in the definition of $RF(X)$ given in the question.] $\endgroup$
    – Pierre-Yves Gaillard
    Commented Mar 21, 2017 at 19:09
  • $\begingroup$ @Pierre-YvesGaillard, Since we can identify $F(-)$ with ${\rm Hom}_{{\cal C}}({\bf Z}/2,-)$ and since $X$ can be identified with the product of ${\bf Z}/2[n]$ we get that $R_0 F(X) \cong \prod_{n \in {\bf Z}} R_0{\rm Hom}({\bf Z/2},{\bf Z/2}[n]) \cong \prod_{n \in {\bf Z}} R_0{\rm Ext^{-n}}({\bf Z/2},{\bf Z/2}) \cong \prod_{n \in {\bf Z}} {\bf Z}/2$. $\endgroup$
    – Yonatan Harpaz
    Commented Mar 21, 2017 at 19:11
  • $\begingroup$ @Pierre-YvesGaillard, there are several approached to derived functors. I admit I am less familiar with the definition via universal colimits, but on chain complexes of (not necessarily finite dimensional) vector spaces there are also various model structures which one can use to compute derived functors. The definition with universal colimits seems a bit hard to verify in general (I suppose in the book they give some useful criteria). $\endgroup$
    – Yonatan Harpaz
    Commented Mar 21, 2017 at 19:16
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    $\begingroup$ Actually the formula in my previous comment is inaccurate. There are only ${\bf N}$ worth of copies of ${\bf Z}/2$ in the end, not ${\bf Z}$ worth. Also I may have confused the positive and negative grading conventions. $\endgroup$
    – Yonatan Harpaz
    Commented Mar 21, 2017 at 19:23

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