# Universal homeomorphism of stacks and etale sites

A morphism between schemes is a universal homeomorphism if it is integral, surjective, universally injective. For morphism between algebraic stacks, this notion also make sense.

It is well know that a universal homeomorphism between schemes induces:

(1) homeomorphism between their underlying topological spaces; and

(2) equivalence between their etale sites.

I have seen the proof of these statements some time ago but didn't really understand well. So first a somewhat vague question:

Question 1: Is there a good conceptual reason why for schemes, homeomorphism between underlying topological spaces should imply equivalence between etale sites?

For me, at least psychologically, the underlying topological space and the etale site of a scheme seems quite unrelated. So I would prefer an intuitive explanation of this (seemingly?) coincidence.

Now for universal homeomorphism of stacks, one can also define its underlying topological space and its etale site and as far as I know, (1) is still true. It seems (2) may not be true, but something weaker is true, i.e. it induces equivalence of etale topoi, see for example Remark 4.26 of Behrend's book here.

In particular, if $\mathcal{X}$ is a (reasonable) Deligne-Mumford stack, the Keel Mori theorem says that it has a coarse moduli space $X$ and the natural morphism $\mathcal{X}\to X$ is a proper universal homeomorphism. (see the second paragraph in Conrad's paper).

Thus the etale topos of $\mathcal{X}$ is isomorphic to the etale topos of $X$. I'm trying to understand this isomorphism in the following special case:

Let a finite group $G$ acts on a ring $A$. Then my understanding is that the stack quotient $[(Spec A)/G]$ has a coarse moduli space $Spec(A^G)$, i.e. the GIT quotient. Then the etale topos of $[(Spec A)/G]$ is isomorphic to that of $Spec(A^G)$. In particular, when $A$ is an algebraically closed field with trivial $G$ action, then I reached a suspicious conclusion that the etale topos of $BG$ is isomorphic to etale topos of a point.

Question 2: Is this correct? If yes, what is the key (commutative algebra) ingredient behind this isomorphism of topos in this special case? Otherwise any comments on where I made mistakes are welcome.

• Something is wrong with your analysis in Question 2. For instance it is clear that if $G$ acts trivially on $A=k$ a field, then $BG$ and $spec(k)$ don't have equivalent étale sites. – Niels Jul 28 '17 at 7:42
• @Niels: I'm also confused by this. What about their etale topos? I guess in general inequivalent sites could give rise to equivalent topos? – Jingren Chi Jul 28 '17 at 15:00
• Objects in the etale site of $BG$ can have more automorphisms than the corresponding objects of the etale site of $\text{Spec}(k)$. – Jason Starr Jul 28 '17 at 16:34
• @HCCH : if $k$ is algebraically closed, the étale topos of $BG$ is $G$-Sets, so it is not equivalent to the étale topos of $spec (k)$, which is Sets. So considering the topoi doesn't help. – Niels Jul 28 '17 at 20:08

(1) is by definition. The standard proof of (2) is via descent of étale morphisms along universal submersions (see SGA1, Exp IX, Thm 4.10, or http://stacks.math.columbia.edu/tag/04DY, or my paper "Submersions and effective descent of étale morphisms", Thm 5.21). The point is that given a universal homeomorphism $f\colon X'\to X$ and an étale morphism $E'\to X'$, it comes equipped with a unique descent datum along $f$ because the diagonal of $f$ is a nil-immersion. Since $f$ is a universal submersion, the descent datum is effective: $E'\to X'$ is the pull-back of an étale morphism $E\to X$.

I believe that one could prove that we have an equivalences of étale topoi without this result. Namely, it is enough to prove that the unit and counit of the adjunction $(f^*,f_*)$ are natural isomorphisms which is easy using the description of $f_*$ of a finite (or integral) morphism. This uses the description of finite morphisms over strictly henselian local rings.

Concerning Question 1, the site-equivalence is more subtle than the topoi-equivalence, essentially due to the difference between schemes and algebraic spaces. The topoi-equivalence is more conceptual but does depend on the properties of finite morphisms over henselian rings as we will see.

The equivalence of étale sites/topoi fails as you have stated it for algebraic spaces and stacks. It is true for a separated homeomorphism of algebraic spaces and hence for representable and separated homeomorphisms of algebraic stacks. It is false for a non-separated homeomorphism of algebraic spaces (e.g., see my paper cited above). It is also false for non-representable separated universal homeomorphisms in general. This highlights that the argument on finite/integral morphisms in the topoi-argument is crucial and hence perhaps not completely "intuitive".

Concerning Question 2, Behrend has a non-standard, stricter, definition of universal homeomorphism (Def 4.18). In that definition, it is not sufficient that $f\colon X\to Y$ is a homeomorphism after arbitrary base change. The definition says that there should be a representable separated universal homeomorphism $X'\to X$ such that $X'\to X\to Y$ is a representable separated universal homeomorphism (Behrend says "schematic" but this is equivalent to separated in this context).

Example: In Behrend's notation: $BG\to *$ is a universal homeomorphism if $G$ is an infinitesimal group (that is, $G\to *$ is a homeomorphism, e.g., $G=\mu_p$ or $G=\alpha_p$ in characteristic $p$) but not if $G$ is étale (e.g., $G=\mathbb{Z}/n\mathbb{Z}$).

Since representable separated universal homeomorphism give rise to equivalences of étale topoi, it is obvious that Behrend's definition does as well.

When it comes to a coarse moduli map $\pi\colon \mathcal{X}\to X$, it is a proper (hence separated) universal homeomorphism, but not representable, and it typically does not give rise to an equivalence of étale sites or topoi (it does for $BG\mu_p\to *$). The correct statement is the following:

Theorem: There is an equivalence of stabilizer-preserving étale morphisms $\mathcal{E}\to \mathcal{X}$ and étale morphisms $E\to X$ (this also holds for étale morphisms that are representable by algebraic spaces, hence also for étale sheaves).

Here stabilizer-preserving can be taken to mean pointwise stabilizer-preserving: for every $e\colon \operatorname{Spec} k\to \mathcal{E}$ with image $x\colon \operatorname{Spec} k\to \mathcal{X}$, the homomorphism $\mathrm{stab}(e)\to \mathrm{stab}(x)$ is an isomorphism of $k$-groups.

You can find a proof (for quasi-separated étale morphisms) in my paper "Existence and properties of geometric quotients", Thm 6.12 (via 6.6-6.7).

• Thanks for your answer and reference! Another question regarding the Theorem you stated: if an etale sheaf $\mathcal{F}$ on $\mathcal{X}$ corresponds to an etale sheaf $\mathcal{G}$ on the coarse moduli $X$, are their cohomology isomorphic? Do you know a reference for the statement about cohomology? – Jingren Chi Aug 15 '17 at 15:00
• Yes, if the stack is tame (inertia is prime to the characteristic) and the sheaf is torsion. Since $\mathcal{G} = \pi_* \mathcal{F}$ it is enough to prove that $\pi_*$ has no higher cohomology (which also proves that $H^i(\mathcal{X},\mathcal{F})=H^i(X,\pi_*\mathcal{F})$ for any sheaf). By proper base change, we may reduce to $\mathcal{X}=BG$ and then $\pi_*$ is taking $G$-invariants which is exact. See Zheng "Six operations ... for Deligne–Mumford stacks" (2015), Lemma 2.14 / Proposition 2.15 or Abramovich–Corti–Vistoli (2003), Appendix A [this proof apparently doesn't need $F$ torsion]. – David Rydh Aug 16 '17 at 16:32