Given an abelian category $A$, we can form its derived category $D(A)$ as follows:
- Start with the category of (co)chain complexes in $A$.
- Construct the homotopy category $K(A)$ by keeping the objects, but defining the morphisms by quotienting by (co)chain homotopy.
- Localize with respect to quasi-isomorphisms, which admits both-sided calculus of fractions.
Therefore, given any (co)chain complexes $X$ and $Y$ in $A$, we have:
$$\operatorname{Hom}_{D(A)}(X,Y) = \varinjlim_{s: Y \to Y'} \operatorname{Hom}_{K(A)}(X, Y')$$
where the limit is taken over all quasi-isomorphisms $s: Y \to Y'$. (Gabriel–Zisman, Calculus of fractions and homotopy theory, Proposition I.2.4)
My question is: Can the filtered colimit always be taken to be a set?
Best Answer
No. Let $R=k[x_i\mid i\in I]$ be a polynomial "algebra" in variables indexed by a proper class $I$. [I put "algebra" in quotation marks, as of course $R$ is not a set.]
Let $A$ be the category of finite dimensional $R$-modules (i.e., finite dimensional vector spaces with a class of commuting endomorphisms indexed by $I$). This is a perfectly good locally small abelian category.
Let $k$ be the one-dimensional $R$-module on which every $x_i$ acts as zero.
For each $i\in I$, there is an extension of $k$ by $k$ on which $x_i$ acts as $\begin{pmatrix}0&1\\0&0\end{pmatrix}$ and on which $x_j$ acts as zero for $j\neq i$. These extensions are clearly pairwise nonequivalent, and so $$\operatorname{Hom}_{D(A)}(k,k[1])=\operatorname{Ext}^1_R(k,k)$$ is not a set.