# How much is a topological space $X$ determined by the category of sheaves of abelian groups on $X$?

Well, the title pretty much says it all. We have a functor $$\mathsf{Sch}_{Ab} : \mathsf{Top} \to \mathsf{Cat}$$ which takes a topological space $X$ to the category $\mathsf{Sch}_{Ab}(X)$ of sheaves of abelian groups on $X$. Every continuous map $f:X \to Y$ gives rise to a functor $f_*:\mathsf{Sch}_{Ab}(X) \to \mathsf{Sch}_{Ab}(Y)$, which takes a sheaf $\mathcal{F}$ on $X$ to the direct image sheaf $f_*\mathcal{F}$ on $Y$.

If $\mathsf{Sch}_{Ab}(X)$ and $\mathsf{Sch}_{Ab}(Y)$ are isomorphic, or say equivalent, can we say anything in turn about $X$ and $Y$? I suppose not...

On a similar note: we can also consider $\mathsf{Sch}_{C}$ as a functor of $C$, and then we can see that $\mathsf{Sch}_{C} \cong \mathsf{Sch}_{D}$ implies $C \cong D$, since, for the one-point space $*$, $\mathsf{Sch}_{C}(*) \cong C$.

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Would you consider using \mathsf and \mathcal, instead of \mathfrak and \mathscr, for categories and sheaves respectively? I find them to be far more readable. – Zev Chonoles May 8 '12 at 5:56
@ZevChonoles, done! I personally prefer the romantic notation of Hartshorne but I can definitely see where you're coming from. – Bruno Joyal May 8 '12 at 6:04
To write down something pretty obvious: If you don't put conditions on $X$ and $Y$ you can certainly not conclude that $X$ and $Y$ are homeomorphic if $\mathsf{Sch}_{\mathsf{Ab}}(X)$ and $\mathsf{Sch}_{\mathsf{Ab}}(Y)$ are equivalent. Take two finite spaces with the trivial topology, that have different cardinality. [I think my point is that if you can say something, you will say it about $\mathsf{Open}(X)$ and $\mathsf{Open}(Y)$, since sheaves do not see points, but only opens.] – jmc May 8 '12 at 7:01
That's a good point, @jmc. – Bruno Joyal May 8 '12 at 20:24

It is well-known, at least amongst topos theorists, that the pseudofunctor $\textbf{Sh} : \textbf{Top} \to \mathfrak{BTop}$ factors through the category of locales $\textbf{Loc}$, and the pseudofunctor $\textbf{Sh} : \textbf{Loc} \to \mathfrak{BTop}$ is essentially full and faithful.

Indeed, first, suppose $\mathcal{E} = \textbf{Sh}(X)$, where $X$ is a topological space. Then, the frame of subobjects of the terminal object of $\mathcal{E}$ is exactly the frame of open subsets of $X$. On the other hand, it is clear from the definition that $\textbf{Sh}(X)$ only depends on the frame of open subsets of $X$ and not on the points, so two topological spaces which have isomorphic frames will have isomorphic toposes – and two spaces with equivalent toposes will have isomorphic frames.

Now, suppose $\mathcal{E}$ and $\mathcal{F}$ are localic toposes, meaning of the form $\textbf{Sh}(X)$ for some locale $X$. A geometric morphism $f : \mathcal{E} \to \mathcal{F}$ is completely determined by its inverse image part $f^* : \mathcal{F} \to \mathcal{E}$, which is by definition left exact and cocontinuous. But this means that $f^*$ restricts to a frame homomorphism from the frame of subterminals of $\mathcal{F}$ to the frame of subterminals of $\mathcal{E}$ – which is precisely the data of a locale morphism $\operatorname{Sub}_\mathcal{E} 1 \to \operatorname{Sub}_\mathcal{F} 1$ – and because a localic topos is generated by its subterminals, this completely determines $f$ as a geometric morphism.

Finally, since you say you are reading Hartshorne, I'll mention this: the underlying topological space of a scheme is a sober topological space, so two schemes are homeomorphic if and only if their frames of open sets are isomorphic. Moreover, given only the frame of open sets, one can recover the topological space by looking at frame homomorphisms $O \to \{ 0, 1 \}$.

What I said above is really about sheaves of sets. But I'm sure that something similar can be said about the sheaves of abelian groups, albeit with more effort: after all, one can apply the free group functor to the subterminals to obtain a family of abelian sheaves. The trouble is figuring out what is special about these sheaves... but here is one possibility: we take $\textbf{Ab}(\textbf{Sh}(X))$ as a symmetric monoidal closed category. The free group functor $\textbf{Sh}(X) \to \textbf{Ab}(\textbf{Sh}(X))$ is strong monoidal (if I'm not mistaken), preserves monomorphisms and is cocontinuous, so we expect to see the frame of open sets reflected in some way in $\textbf{Ab}(\textbf{Sh}(X))$. I believe it is precisely the frame of subobjects $A$ of the constant sheaf $\mathbb{Z}$ such that $A \otimes_\mathbb{Z} A \cong A$.

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Thanks for your great answer Zhen. I'll try to wrap my head around it (but I think I'm missing some background). Would you have reading recommendations in topos theory? – Bruno Joyal May 8 '12 at 20:25
@Bruno: Mac Lane and Moerdijk's Sheaves in geometry and logic is not too bad, and after that there's Johnstone's Sketches of an elephant (though, as a reference more than a textbook). – Zhen Lin May 8 '12 at 21:36
Thank you @ZhenLin! – Bruno Joyal May 9 '12 at 3:08