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 $\text D(\mathcal C')$.


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.

  • $\begingroup$ Thanks! Don't you mean ${\rm Ext}^n_{\cal C}$ instead of ${\rm Ext}^n_{\cal C'}$? Don't you have ${\rm Ext}^n_{\cal C'}({\bf Z}/2,{\bf Z}/2)\cong0$ for all $n\neq0$? $\endgroup$ – Pierre-Yves Gaillard Mar 21 '17 at 10:27
  • $\begingroup$ 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 Mar 21 '17 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 Mar 21 '17 at 19:09
  • $\begingroup$ @Pierre-YvesGaillard, you're right of course, it should be ${\cal C}$, I fixed it now. 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 Mar 21 '17 at 19:11
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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 Mar 21 '17 at 19:23

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