Let $C$ be a category and let $F:C\rightarrow D$ be a functor with $D$ locally presentable and cartesian closed. When does the Yoneda extension $\widehat{F}=Lan_{y} F:[C^{op},Set]\rightarrow D$ preserve finite products?

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    $\begingroup$ I think the answer is "iff the category of elements of $F$ is sifted". I won't be able to write this up as an answer for a while, but hopefully an answer will appear soon, even if not by me. $\endgroup$ – Todd Trimble Nov 22 '16 at 1:53
  • $\begingroup$ @ToddTrimble But in my case the codomain category of $F$ is not necessarily $Set$, so how does it makes sense to talk about its category of elements? $\endgroup$ – user84563 Nov 22 '16 at 2:25
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    $\begingroup$ Well, you've got a point there. $\endgroup$ – Todd Trimble Nov 22 '16 at 2:36
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    $\begingroup$ In that case you should probably ask for the categories of elements of $D(G_i, F -)$ to be sifted for some chosen generating family $(G_i)$. $\endgroup$ – Karol Szumiło Nov 22 '16 at 10:10
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    $\begingroup$ Karol's answer seems to imply that there is a relation between $\hat F$ preserving products and the siftedness of the categories of elements of the profunctors $\varphi_F$; I'll try to dig into that since I find it interesting per se (I suggest, if you want to go that way, to look at Thomas Streicher's note on Bénabou "distributors at work"). For the time being I also think that a sufficient condition for what you want is that $\hat F$ commutes with finite products of representables $\endgroup$ – Fosco Nov 23 '16 at 10:28

(Update: the following answer applies only in the case that $C$ has finite products.)

The left Kan extension $\hat{F}$ preserves finite products just when $F$ does.

One direction is easy since $F \cong \hat{F} \circ Y$ and $Y$ preserves finite products.

The converse is an old result of Borceux and Day from their paper On finite product preserving Kan extensions. in the Bulletin of the Australian Mathematical Society. Combining Theorem 1.5 with Example 3.1 of that paper gives the result.

(Warning: they do everything in maximal generality and in the enriched setting so their results may take some time to parse.)

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    $\begingroup$ True, but then again the OP doesn't assume $C$ has products. $\endgroup$ – Todd Trimble Nov 25 '16 at 4:55
  • $\begingroup$ Oh right - somehow I didn't spot that. $\endgroup$ – john Nov 25 '16 at 5:43
  • $\begingroup$ @john Does the same hold if we replace "finite products" with "finite limits?" $\endgroup$ – user84563 Dec 11 '16 at 16:30
  • $\begingroup$ In answer to the last question: not quite. For a simple example, let $C = D$ be a complete Heyting algebra, and take $F: C \to D$ to be the identity. Then $\hat{F}: Set^{C^{op}} \to C$ is the reflector that is left adjoint to the Yoneda embedding $y: C \to Set^{C^{op}}$, but $\hat{F}$ is not left exact (otherwise $C$ would be a Grothendieck topos). If however $D$ is a Grothendieck topos, then the answer is yes; this is a corollary of Diaconescu's theorem. $\endgroup$ – Todd Trimble Dec 26 '16 at 14:18

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