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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.)

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  • $\begingroup$ 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. $\endgroup$ – Zhen Lin Jul 22 '15 at 9:45
  • $\begingroup$ @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... $\endgroup$ – Pece Jul 22 '15 at 12:45
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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.

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