# Left Kan extension of a $\mathsf{Set}$-valued finite-product-preserving functor

I've been told that the following is true:

Proposition. Consider $\mathcal A,\mathcal B$ small categories with finite products and $j\colon \mathcal A \to \mathcal B$ preserving them. Then for any finite-product-preserving $F \colon \mathcal A \to \mathsf{Set}$, its left Kan extension $j_!F$ also is finite-product-preserving.

And I'm inclined to believe it because of the case where $\mathcal A = \mathsf{FinSet}$ and $j$ is bijective-on-objects: it gives that the forgetful functor for the Lawvere theory $j$ admits a left adjoint (the free functor).

Here's my attemp at proving the proposition:

1. Remark that for any $a \in \mathcal A$, the left Kan extension $j_!{\mathcal A(a,-)}$ is just $\mathcal B(j(a),-)$.
2. For $F \colon \mathcal A \to \mathsf{Set}$ considered as a preasheaf on $\mathcal A^\circ$ (opposite category of $\mathcal A$), one has $$F = \operatorname{colim}\left( \mathcal A^\circ / F \to \mathcal A^\circ \stackrel h \to [\mathcal A,\mathsf{Set}] \right)$$ where $h$ is the Yoneda embedding.
3. Left adjoints commute with colimits, so $$j_!F = \operatorname{colim}_{f:h(a)\to F}\left( \mathcal B(j(a),-) \right)$$
4. Functors of the form $\mathcal B(b,-)$ preserves finite products by definition, and filtered colimits commute with finite limits in $\mathsf{Set}$. So we conclude by showing that $\mathcal A^\circ / F$ is filtered.
5. As $F$ preserves binary products, any diagram over the discrete category with two objects admits a cocone.

But what about the diagram over $\bullet \rightrightarrows \bullet$ ?

I don't see why those should admit a cocone... I didn't yet take advantage of the fact that the terminal of $\mathcal A$ is mapped by $F$ to a singleton, but it does not seem to help. Also, I did not use the finite-product-preserving hypothesis on $j$, which I believe is necessary.

Can someone help me finish/correct this sketch of proof ? (Or provide a counter-example if the proposition is false.)

• I don't think you will be able to show that the category of elements in question is filtered. At any rate, it would suffice to show that it is sifted. Jul 22, 2015 at 9:45
• @ZhenLin Of course, sifted is enough, tanks. As $F$ preserves finite products, the category $\mathcal A^\circ / F$ has finite coproducts, so is sifted. So the hypothesis on $j$ seems superfluous...
– Pece
Jul 22, 2015 at 12:45

You can prove the statement without appealing to the sifted VS discrete commutation between limits and colimits; it's a far more general result about promonoidal functors.

This result tells you that if you are given a strong promonoidal functor $j\colon \mathcal{A}\to \mathcal{B}$ between promonoidal categories, then the left Kan extension $\text{Lan}_j\colon [\mathcal{A},\mathbf{Set}]\to [\mathcal{B},\mathbf{Set}]$ is a strong monoidal functor, when the presheaf-categories are endowed with the convolution product $$F\star G = \int^{A', A''} \mathbf{P}(A', A'',-)\times FA' \times GA''.$$ This boils down to your statement choosing the "trivial" promonoidal structure $\mathbf{P}(A', A'', X) = \hom(A', X)\times \hom(A'',X)$, as in that case $(F\star G)(X)\cong FX \times GX$.[1]

A promonoidal functor between promonoidal categories $h : (\mathcal A, P_{\cal A}, J_{\cal A})\to (\mathcal B, P_{\cal B}, J_{\cal B})$ is a functor $h\colon \mathcal A\to \mathcal B$ endowed with natural transformations $$\begin{gather} P_{\cal A}(A,A'; A'') \to P_{\cal B}(hA,hA'; hA'') \\ J_{\cal A}A \to J_{\cal B}(hA) \end{gather}$$ which induce natural isomorphisms $$\begin{gather} \int^{A''}P_{\cal A}(A, A';A'')\times \hom(hA'', B)\cong P_{\cal B}(hA, hA'; B) \\ \int^A J_{\cal A}A \times \hom(hA, B)\cong J_{\cal B}B \end{gather}$$

Here's the coend rollercoaster (the trick is that $\text{Lan}_UV$ is the coend $\int^X \hom(UX,-)\times VX$)[2]:

\begin{align} \text{Lan}_j(F\star G)(B) &\cong \int^A \mathcal{B}^{jA}_B \times (F\star G)_A\\ (1)&\cong \int^A \mathcal{B}^{jA}_B \times \left[ \int^{A', A''}P_A^{A'A''}\times F_{A'}\times G_{A''}\right]\\ (2)&\cong \int^{AA'A''}\mathcal{B}^{jA}_B \times P_A^{A'A''}\times F_{A'}\times G_{A''}\\ (3)&\cong \int^{A'A''} P_B^{jA'jA''}\times F_{A'}\times G_{A''}\\ (4)&\cong \int^{A'A''B'B''}\mathcal{B}^{jA'}_{B'}\times \mathcal{B}^{jA''}_{B''}\times P_B^{B',B''}\times F_{A'}\times G_{A''}\\ (5)&\cong \int^{B'B''}P(B',B'',B)\times \left[\int^{A'} \mathcal{B}^{jA'}_{B'}\times F_{A'}\right ]\times \left[\int^{A''} \mathcal{B}^{jA''}_{B''}\times G_{A''}\right ]\\ &\cong (\text{Lan}_j F\star \text{Lan}_j G)(B) \end{align}

(1) is true by definition; (2) is true since $\bf Set$ is cartesian closed and products distribute over colimits; (3) is true since the functor $j$ is strong promonoidal; (4) is true, it's a form of Yoneda expansion/reduction; to obtain (5) you only have to reorder factors.

[1] Do not worry, I'm not assuming you're comfortable with the definition of promonoidal category; the idea behind them is that a promonoidal category is what you obtain when you take the definition of a monoidal category and you replace every occurrence of the word functor with the word profunctor. Appendix A here contains a basic introduction to the basic definitions.

[2] I adopt "Einstein notation" [coends, Def. 5.3]: superscripts are contravariant and subscripts are covariant components of functors. Co-ending is always on repeated indices, one covariant and one contravariant. There's no particular reason for it, but the fact that equations are cleaner and do not flood the page.