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Let me first recap the definitions of flatness, using the nLab's terminology:

A functor into $F : \mathcal{C} \to \mathbf{Sets}$ is Set-valued flat when the category of elements $\int E$ is filtered. This implies, and is implied by, left exactness of the Kan extension $\mathrm{Lan}_{よ}(F) : [\mathcal{C}^{\mathrm{op}}, \mathbf{Sets}]$. A functor $F : \mathcal{C} \to \mathcal{E}$ is representably flat when each $\mathrm{Hom}_E(B, F(-)) : \mathcal{C} \to \mathbf{Sets}$, for $B$ ranging over the objects of $\mathcal{E}$, is Set-valued flat.

The nLab claims that for a representably flat functor $F : \mathcal{C} \to \mathcal{E}$, where both categories are small, the extension of $F$ to a functor $[\mathcal{C}^{\mathrm{op}}, \mathbf{Sets}] \to [\mathcal{E}^{\mathrm{op}}, \mathbf{Sets}]$ preserves finite limits. I'm interested in a different setup: If $\mathcal{E}$ is finitely complete and cocomplete, does representable flatness of $F : \mathcal{C} \to \mathcal{E}$ ensure that $\mathrm{Lan}_{よ}(F) : [\mathcal{C}^\mathrm{op}, \mathbf{Sets}] \to \mathcal{E}$ is left exact? If this is not the case for an arbitrary cocomplete category $\mathcal{E}$, does the answer change when $\mathcal{E}$ is a sheaf topos?

Amélia Liao
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    Corollary 3.3 here https://math.au.dk/~kock/postulated.pdf seems to answer "yes" to the very same question, when $\cal E$ is a Grothendieck topos – fosco Apr 24 '22 at 08:55
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    This is not true in general because it would imply for instance that any finite lattice is distributive. Take any non-distributive finite lattice $L$ seen as a small category. Then the identity $L→L$ is representably flat but $\hat{L}→L$ cannot be left exact because then $L$ would be a quotient of a frame. In the context of posets instead of categories, frames are the cocomplete posets making your question true... So something similar might be true for categories. But I don't know if the definition of flatness that needs to be considered is representable flatness or something stronger... – Dabouliplop Apr 24 '22 at 12:32
  • I just want to mention that representable flatness implies internal flatness (the converse is false). I never saw this mentioned anywhere (despite searching; maybe people consider this too trivial to mention? I don't understand why I found this said nowhere, maybe I missed it). When I said "something stronger"... actually it was something weaker. – Dabouliplop Apr 25 '22 at 16:07
  • The proof: take $F:C→E$ representably flat. Take a finite diagram $(X_i)_i$ in $C$. Applying representable flatness to $\lim_i F(X_i)$, the cone $\lim_i F(X_i)→F(X_i)$ factors through some $F(Y)$, and $F(Y)→\lim_i F(X_i)$ is epi because it has a section. – Dabouliplop Apr 25 '22 at 16:07

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I assume you mean that $\mathcal E$ is not only cocomplete, but also finitely complete (otherwise I'm not sure what you would mean by a left-exact functor with codomain $\mathcal E$). Then Garner and Lack's paper Lex Colimits seems to answer part of your question ; proposition 2.4(5) and 2.5 combined assert that :

Given a cocomplete and finitely complete category $\mathcal E$, the following are equivalent :

  • for every lex functor $F : \mathcal K \to \mathcal E$ (with $\mathcal K$ finitely complete), its left extension $\mathrm{Lan}_{よ}F : [\mathcal K^{op}, \mathrm{Set}] \to \mathcal E$ is left exact
  • $\mathcal E$ is an infinitary pretopos

Given that any lex functor $\mathcal D \to \mathcal E$ is representably flat, it follows that your $\mathcal E$ needs to be an infinitary pretopos at least.

Elies
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  • There is a school of thought that defines "left exact functor" = "representably flat" when the codomain category is not assumed to have finite limits. – Zhen Lin Apr 25 '22 at 01:13
  • I did mean to mention E is assumed lex, I'll update the question. – Amélia Liao Apr 25 '22 at 12:53