Fix a complete, cocomplete, symmetric monoidal closed category $\mathcal{V}$. I will also assume that there is a forgetful functor $\mathcal{V}\to \mathbf{Set}$ with a left adjoint. By standard results, this adjunction lifts to a 2-adjunction between the 2-category of categories and the 2-category of $\mathcal{V}$-categories; IOW, we can speak of the free $\mathcal{V}$-category on a category.

Now, let $\mathcal{A}$ be a (small) symmetric monoidal closed $\mathcal{V}$-category. Then the $\mathcal{V}$-presheaf category $\mathbf{PShv}(\mathcal{A})$ is symmetric monoidal closed when endowed with the Day convolution structure and the Yoneda embedding is symmetric monoidal closed. Furthermore, any symmetric monoidal $\mathcal{V}$-functor $F:\mathcal{A}\to \mathcal{B}$ with $\mathcal{B}$ cocomplete, lifts uniquely (up to unique isomorphism) to a cocontinuous symmetric monoidal functor $\mathbf{PShv}(\mathcal{A})\to \mathcal{B}$ by taking the left Kan extension $L_{Y}(F)$ of $F$ along Yoneda $Y$.

I can prove all this to myself without great effort. What has left me stumped is the following question:

Q: assume $\mathcal{B}$ and $F$ closed, is $L_{Y}(F)$ closed?

Since Yoneda is closed the converse is trivially true. If the answer is negative in general, is there any criteria to have a positive answer? If it helps the answer, my use case is when $\mathcal{A}$ is the free $\mathcal{V}$-category on a Heyting algebra (e.g. the open sets of a topological space) with monoidal structure the intersection. Since in a lattice there is at most one arrow between two objects, Day convolution degenerates into the pointwise tensor product.

Thanks in advance, regards, G. Rodrigues


2 Answers 2


There are various meanings of "closed functor", and the answer will depend on which meaning is adopted.

The meaning that I imagine Buschi Sergio adopted was what might be called "lax closed", where the structural constraint on a lax closed functor $F$ is a map $F(x \Rightarrow y) \to F(x) \Rightarrow F(y)$ satisfying some naturality and coherence conditions. Such a structure comes for free if $F$ is a (lax) monoidal functor, and since the left Kan extension of a lax monoidal functor is again lax monoidal, BS's remark would apply.

But perhaps you meant "strong" (or pseudo) closed, where the structural constraint is an isomorphism. In fact, I'm going to guess that's what you meant, since the other case (just discussed) is more or less obvious.

Since the construction of the internal hom (adjoint to the Day convolution monoidal product) on the presheaf category will inevitable involve limits, and since the left Kan extension does not generally preserve limits, the odds are highly against the left Kan being a closed functor in this strong sense. So the strategy should be to test it against just about the simplest case you can think of, and expect something will go wrong. Indeed, let's take $A$ to be the cartesian closed category $2$ (with two objects $0, 1$ and exactly one non-identity arrow, running in the direction $0 \to 1$). And take $B$ to be $Set$. Take $F: 2 \to Set$ to be the obvious strong closed functor, taking the object $0$ to the empty set and $1$ to the terminal set.

The following assertions are pretty easy to verify:

(1) The Day convolution on $Set^{2^{op}}$ is cartesian product. Thus the internal hom is the usual hom for the cartesian closed structure on the presheaf category.

(2) The left Kan extension $L_{yon} F$ is easily calculated: $L_{yon}(F)(X) = X(1)$.

(3) $L_{yon}(F)(X \Rightarrow Y)$ is accordingly $(X \Rightarrow Y)(1)$, which one may calculate to be the end

$$\int_{a \in 2} Y(a)^{X(a)}$$

or more explicitly, the pullback of the evident diagram

$$Y(0)^{X(0)} \to Y(0)^{X(1)} \leftarrow Y(1)^{X(1)}$$

(4) But $L_{yon}(F)(X) \Rightarrow L_{yon}(F)(Y)$ in $Set$ is just $Y(1)^{X(1)}$.

Since the results of (3) and (4) differ, we see that $L_{yon}$ is not strong closed.

  • $\begingroup$ I like "the strategy should be to test it against just about the simplest case you can think of, and expect something will go wrong." (-: $\endgroup$ Nov 10, 2010 at 2:51
  • $\begingroup$ Yes. I meant "strong". The lax case is easily disposed of and not very interesting for what I am pursuing. And I agree with Mike Shulman, great strategy! One of those knowledge tidbits that needs to be pointed out before we realize how "obvious" it is. $\endgroup$ Nov 10, 2010 at 11:12
  • $\begingroup$ Well, the phrasing of that strategy is heavily loaded, since you have to have the intuition first of whether it's likely to be right or wrong! But I've often found that if it's a very general statement and you can't prove it by the usual means, there's a good chance that that's because it's wrong! And then maybe here the strategy kicks in. $\endgroup$
    – Todd Trimble
    Nov 10, 2010 at 12:13

See Eilenberg, S. & Kelly, G.M. Closed categories Proceedings of the Conference on Categorical Algebra. (La Jolla, 1965) Springer. 1966

At p. 487 Propositio.43:

Let $V$ and $V'$ monoidal closed categories, the a "monoidal functor data" is identified with a "closed functor data".

In short: a monoidal funtor is closed too.

Then if you (left) Kan extention is (a principal part of) a monoidal functor between monoidal-closed categories, then its also close.

I seems this can work...


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.