Let $\mathcal{M}$ be a complete and cocomplete closed symmetric monoidal category, with unit $I.$

Does the functor $\text{Hom}_{\mathcal{M}}(I,-)$ always preserve filtered colimits?

A reference or a proof is the same for me. I believe this should be true as it's true in many examples such as $\mathbf{Sets},$ categories of modules, presheaf toposes etc.

  • $\begingroup$ Do you mean the internal hom or the Set-valued hom? $\endgroup$
    – Zhen Lin
    Jan 5, 2022 at 22:31
  • $\begingroup$ @Zhen Lin hi. I mean the Set-valued Hom, as in the definition of compact/small object. $\endgroup$
    – user1011965
    Jan 5, 2022 at 22:50

1 Answer 1


It's not even true for all presheaf toposes.

Let $I$ be a set. The category $\textbf{Set}^I$ is a presheaf topos, and in particular, complete and cocomplete and cartesian closed. The terminal object $1$ is not, in general, finitely presentable. Indeed, a general object $X$ in $\textbf{Set}^I$ is just an $I$-indexed family of sets, and the set of morphisms $1 \to X$ is in natural bijection with $\prod_{i \in I} X_i$.

Proposition. The functor $X \mapsto \prod_{i \in I} X_i$ preserves filtered colimits if and only if $I$ is a finite set.

Proof. The "if" direction is well known.

For the "only if" direction, observe that $X \cong \varinjlim_{J \in \mathscr{P}_\textrm{f} (I)} X_{(J)}$, where $\mathscr{P}_\textrm{f} (I)$ is the set of finite subsets of $I$ and $(X_{(J)})_i = X_i$ if $i \in J$ and $(X_{(J)})_i = \emptyset$ if $i \notin J$. If $I$ is infinite, then $\prod_{i \in I} (X_{(J)})_i$ is always empty, hence $\varinjlim_J \prod_{i \in I} (X_{(J)})_i$ is also empty. But $\prod_{i \in I} X_i$ may be non-empty. So $X \mapsto \prod_{i \in I} X_i$ does not preserve filtered colimits if $I$ is infinite. ◼


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