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Let $(\mathcal{C},\mathcal{I})$ and $(\mathcal{D},\mathcal{J})$ be sites where $\mathcal{C}, \mathcal{D}$ are categories and $\mathcal{I}$ and $\mathcal{J}$ are Grothendieck topologies on $\mathcal{C}$ and $\mathcal{D}$ respectively.

I am trying to understand what it means to say a morphism of sites $(\mathcal{C},\mathcal{I})\rightarrow (\mathcal{D},\mathcal{J})$.

This should be atleast a functor $F:\mathcal{C}\rightarrow \mathcal{D}$.

As Grothendieck topologies comes with notion of covering, it is reasonable to ask that this functor $F$ is compatible with notion of covering. That is, for each covering $\{U_i\rightarrow U\}$ of $U$ in $\mathcal{C}$ the collection $\{F(U_i)\rightarrow F(U\})$ is a covering of $F(U)$ in $\mathcal{D}$.

As there is a condition on pullback, that is, for each covering $\{U_i\rightarrow U\}$ for $U$ in $\mathcal{C}$ and a morphism $V\rightarrow U$ in $\mathcal{C}$, the pull-back $\{U_i\times_U V\rightarrow V\}$ is a covering for $V$ in $\mathcal{C}$, we can ask that, $F$ is compatible with this notionof pullback, that is, for each covering $\{U_i\rightarrow U\}$ of $U$ in $\mathcal{C}$ and a morphism $V\rightarrow U$ in $\mathcal{C}$ we have $F(U_i\times_UV)\cong F(U_i)\times_{F(U)}F(V)$. As we have already asked that $\{F(U_i)\rightarrow F(U)\}$ to be a covering for $F(U)$, it will automatically follow that $\{F(U_i)\times_{F(U)}F(V)\rightarrow F(V)\}$ is a covering for $F(V)$.

This is what I think a morphism of sites should be like.

n-lab page on Morphism of sites asks that the functor is "covering flat". I think it is more than what I have asked above. I only know what is flat functor but could not track what is covering flat. I see that every flat functor has the property that I have mentioned above, but it is more than that.

So, Is my definition of morphism of sites not correct/reasonable?

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Short answer: A morphism of sites as you define it (a functor which preserves covers and the fibre products showing up in the gluing condition) gives rise to an adjunction $$f_s:\mathrm{Sh}(\mathbf{C}) \leftrightarrows \mathrm{Sh}(\mathbf{D}):f^s$$ of sheaf categories. The condition of covering-flatness ensures that $f_s$ is exact. (The resulting structure is known as a geometric morphism of topoi.)


EDIT to clarify subscript/superscript Notation: In the classic example of a map $f:X \to Y$ inducing a morphism of sites $f^{-1}:\mathrm{Op}(Y) \to \mathrm{Op}(X)$, the functors above are the direct image functor $${(f^{-1})}^s = f_\ast$$ and the inverse image functor $${(f^{-1})}_s = f^\ast.$$ ("The ${(-)}^{-1}$ exchanges subscripts and superscripts.") This notation is commonly used in the general case. The adjunction $(f_s \dashv f^s)$ is then written instead as $(f^\ast \dashv f_\ast)$.


Long answer: A functor of small categories $$f:\mathbf{C} \to \mathbf{D}$$ gives rise to an adjoint pair of functors between presheaf categories $$f_p:\mathrm{PSh}(\mathbf{C}) \leftrightarrows \mathrm{PSh}(\mathbf{D}):f^p.$$ The functor $f^p:\mathrm{PSh}(\mathbf{D}) \leftrightarrows \mathrm{PSh}(\mathbf{C})$ is given by composition with $f$. It preserves limits because limits in presheaf categories are computed pointwise. We may build its left adjoint $f_p$ explicitly: it is given by left Kan extension along $f$. The functor $f_p$ preserves finite limits if and only if $f$ is representably flat [nFF, Prop 2.6]. (The left Kan extension is pointwise computed by a colimit. Representable flatness ensures that these colimits are filtered. Finite limits commute with filtered colimits.)

Now assume $\mathbf{C}$ and $\mathbf{D}$ are sites. Sheafification and inclusion over $\mathbf{C}$ form an adjoint pair $$L_\mathbf{C}:\mathrm{PSh}(\mathbf{C}) \leftrightarrows \mathrm{Sh}(\mathbf{C}):I_\mathbf{C}$$ and similarly $(L_\mathbf{D} \dashv I_\mathbf{D})$.

If $f$ preserves covers and the fibre products present in the gluing condition, the functor $f_p$ preserves sheaves: the composite $$f^p \circ I_\mathbf{D}:\mathrm{Sh}(\mathbf{D}) \to \mathrm{PSh}(\mathbf{D}) \to \mathrm{PSh}(\mathbf{C})$$ lands in the subcategory $\mathrm{Sh}(\mathbf{C}) \subset \mathrm{PSh}(\mathbf{C})$. It has a left adjoint $$L_\mathbf{D} \circ f_p: \mathrm{PSh}(\mathbf{C}) \to \mathrm{PSh}(\mathbf{D}) \to \mathrm{Sh}(\mathbf{D}).$$ The adjunction restricts to an adjunction of presheaf categories $$f_s:\mathrm{Sh}(\mathbf{C}) \leftrightarrows \mathrm{Sh}(\mathbf{D}):f^s.$$ The functor $f_s$ is the composite $L_\mathbf{C} \circ f_p \circ I_\mathbf{D}$. We're again looking for a condition that will make this functor exact. As a left adjoint, it already preserves colimits.

Now comes the punchline: Covering-flatness of $f$ is equivalent to the requirement that the composite $L_\mathbf{C} \circ f_p$ preserve finite limits [nFF, Prop 2.15]. (The reason is the same as in the story for presheaves, except the fact we sheafify allows us to use a slightly weaker condition than representable flatness.) This means $f_s$ is exact iff $f$ is covering-flat.

[nFF] https://ncatlab.org/nlab/show/flat+functor

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    $\begingroup$ Your superscript/subscript notation is a little confusing, since people usually write the left adjoint as $f^*$ and the right adjoint as $f_*$. $\endgroup$
    – David Roberts
    Commented Mar 17, 2022 at 23:44
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    $\begingroup$ That's true, thank you! I've added a little paragraph. $\endgroup$ Commented Mar 18, 2022 at 8:34
  • $\begingroup$ @DavidRoberts The notation ($f^p$ for $f_*$ etc) is also used in the Stacks project. I hate it. $\endgroup$ Commented Mar 18, 2022 at 8:53
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    $\begingroup$ @R.vanDobbendeBruyn (Previous comment deleted because I confused myself. I guess that further affirms the point of it being confusing.) I think the superscript/subscript notation indicates that ${(g \circ f)}^p = f^p \circ g^p = $ and ${(g \circ f)}_p = g_p \circ f_p$. The discrepancy in notation may be traced down to the fact that morphisms of sites and maps of topoi have opposite directions. (E.g. in the case of topological spaces, $f:X \to Y$ induces $f^{-1}:\mathrm{Op}(Y) \to \mathrm{Op}(X)$, which induces a map of topoi $\mathrm{Sh}(X) \to \mathrm{Sh}(Y)$.) $\endgroup$ Commented Mar 18, 2022 at 11:14
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    $\begingroup$ Many thanks for taking time to write down an answer for such an old question.. I will not be able to respond immediately.. I will take some time, read it carefully and respond :) $\endgroup$ Commented Mar 18, 2022 at 11:55

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