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Recall that a functor $f_\ast : \mathcal E \to \mathcal F$ between toposes is called a geometric morphism if it has a left exact left adjoint $f^\ast$. Is there an intrinsic characterization of such functors which doesn't refer explicitly to the left adjoint? Let's assume our toposes are Grothendieck.

Of course, the condition that $f_\ast: \mathcal E \to \mathcal F$ admit a left adjoint $f^\ast$ is equivalent, by the adjoint functor theorem, to the condition that $f_\ast$ preserves limits and sufficiently-filtered colimits. But I don't know how to encode the condition that $f^\ast$ preserves finite limits purely in terms of $f_\ast$.

Again by the adjoint functor theorem, one can say when $f^\ast: \mathcal F \to \mathcal E$ is the left adjoint of a geometric morphism $f_\ast$ without referring to $f_\ast$ directly -- it means that $f^\ast$ preserves colimits and finite limits. I'm looking for something similar which refers only to $f_\ast$ and not to $f^\ast$.

I'm also interested in the $\infty$-topos version of this question. It would also be interesting to know the answer when $\mathcal E, \mathcal F$ are Grothendieck abelian categories rather than toposes.

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  • $\begingroup$ Is there some type of object $X$ so that $\hom(-,X)$ takes finite limits to colimits? And are there enough of them to characterize the finite limits? If so, then you are asking that $f_*$ preserves that class of objects. $\endgroup$
    – Theo Johnson-Freyd
    Commented Jan 28, 2020 at 23:31
  • $\begingroup$ Some sort of ind- or pro-thing? $\endgroup$
    – Theo Johnson-Freyd
    Commented Jan 28, 2020 at 23:31
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    $\begingroup$ @TheoJohnson-Freyd Unfortunately no. Suppose that $Hom(\emptyset \times \emptyset, X) = Hom(\emptyset, X) \amalg Hom(\emptyset, X)$. Then since $\emptyset \times \emptyset = \emptyset$ (this is true in any cartesian closed category, for example), we get $2=1$, a contradiction. Here $\emptyset$ is the initial object. There's some sliver of chance such things might exist in the abelian context... $\endgroup$
    – Tim Campion
    Commented Jan 28, 2020 at 23:37
  • $\begingroup$ Small remark: $f^*(1)=1$ means that $\Gamma \circ f_*= \Gamma$, where $\Gamma = \mathrm{Hom}(1,-) : \mathcal{F} \to \mathsf{Set}$ are global sections. $\endgroup$
    – Martin Brandenburg
    Commented Jan 30, 2020 at 3:42
  • $\begingroup$ Also, $f^*$ preserves binary products iff the canonical morphisms $f_*[f^*(B),A] \to [B,f_*(A)]$ are isomorphisms (but here still $f^*$ appears, but at least no products). $\endgroup$
    – Martin Brandenburg
    Commented Jan 30, 2020 at 3:59

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