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Given an operation of say a topological group on a topological space, one can form the quotient stack X//G: the stack associated to the action groupoid. Does this stack satisfy some kind of universal property?

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Probably the simplest example is when the space $X$ is a single point. Then $pt//G$ classifies principal $G$-bundles. Here, the action groupoid is just $G$ considered as a one-object groupoid. The one object, $pt$ becomes an atlas for the stack, so we have a representable epimorphism $pt \to pt//G$. This is in fact the universal principal $G$-bundle! Given any map $X \to pt//G$, the 2-fibered product $X \times_{pt//G} pt\to X$ is a principal $G$-bundle. What we have in general is, given any topological groupoid $H$, its associated stack (stackification of $Hom(blank,H)$ ) is equivalent to $Bun_H:T \mapsto Bun_H(T)$, where $Bun_H(T)$ is the groupoid of principal $H$-bundles over $T$. So, by Yoneda, we have for all spaces $T$, an equivalence of groupoids $Hom(T,Bun_H) \cong Bun_H(T)$. Moreover, you can show that for any map $T \to Bun_H$, the principal bundle $P$ to which the map corresponds is the 2-fibered product $T \times_{Bun_H} H_0 \to T$. Applying this to the case when $H$ is the group $G$, we get what I claimed. Now, if instead $H$ is the action groupoid of $G$ on a space $X$, we get that $X//G$ classifies principal $G \ltimes X$-bundles. This answers the question as far as what maps INTO the stack yield.

As Chris suggested, for any topological groupoid, you can take its enriched-nerve to get a simplicial space. Applying Yoneda on each component, you get a simplicial stack. The weak 2-colimit of this, is equivalent to the stack associated to the groupoid. (In fact, you only need to consider the 2-truncation of this diagram). Spelling this out, you get Proposition 3.19 out of Noohi's Foundations of Topological Stacks I. This roughly says that maps from the stack associated to a groupoid $H$ into another stack $Y$ are the same as maps $f:H_0 \to Y$ (or by Yoneda elements $f \in Y(H_0)_0$) together with an isomorphism $f \circ s \to f \circ t$ satisfying some obvious coherencies, where $s$ and $t$ are the source and target map of the groupoid $H_1 \rightrightarrows H_0$. In the case where $H$ is the action groupoid $G \ltimes X$, you get maps $f:X \to Y$, together with isomorphisms $f \circ proj \to f \circ \rho$, where $\rho:G \times X \to X$ is the action. This becomes more concrete, if for example, $Y$ is not some arbitrary stack, but instead the stack of principal $K$-bundles for some group $K$. Then $Hom(X//G,Bun_K)$ is the groupoid with objects principal $G$-bundles $P$ over $X$ together with an isomorphism $proj^*P \cong \rho^*P$ (satisfying a cocycle condition), with the obvious arrows.

If you were wondering why $X//G$ was the "like the ordinary quotient but better", then the following heuristics should help. The 2-truncated guy whose colimit equals $X//G$ is "like the colimit expressing $X/G$ but with the action built in more". In fact, we we took the UNtruncated guy, so, the simplicial space associated to the action groupoid, then it "IS" the Borel construction (or rather its geometric realization). In particular, this implies the homotopy type of $X//G$ is the same as the homotopy quotient $EG \times_G X$. Now consider one more thing- if $G$ is acting on $X$ we have the commutative square with $G \times X$ on the upper left, $X$ on bottom left and on the upper right, and $X/G$ (the coarse quotient) on the bottom right, with map up top being the action, and the map down below being the projection, then this is a pullback diagram if and only if the action is faithful. If you replace $X/G$ with $X//G$, then this is ALWAYS a (2-)pullback diagram. Hence, going to stacks "makes all actions faithful up to homotopy" (more accurately, stacks allow us to encode the isotropy data that would otherwise be lost in such a way that any action becomes as good as a faithful one).

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My Latex shows up fine in preview, what the heck is going on? –  David Carchedi Jun 15 '10 at 13:51
put mathematical expressions involves asterices in back-quotes. Asterices are used for italics in Markdown, so they confuse things. –  Ben Webster Jun 15 '10 at 14:16
So, I replaced all my asterisks with "pt", but, what do you mean by back-quotes for future reference? –  David Carchedi Jun 15 '10 at 14:31
He refers to the little symbol below the escape key, the key which will give you $\tilde$ if you hold the shift bar. See "How to write math" pointer at the bottom of the right column on some MO screens. You just surround the dollar signs with that extra backtick and then it compiles fine. –  David Jordan Jun 15 '10 at 14:53
Thanks, I found your discussion of pt//G very illuminating! –  Jan Weidner Jun 18 '10 at 14:04

Yes. It is the (homotopy) colimit in stacks of the simplicial diagram:

$$\cdots X \times G^2 \Rightarrow X \times G \rightrightarrows X $$

(here $\Rightarrow$ is a substitute for three arrows). This simplicial object is built from the action of G on X. Since stacks form a 2-category, the natural thing is a "weak" or "homotopy" colimit. See the n-lab pages. The inclusion from spaces into stacks doesn't preserve colimits, so even though this is a diagram of ordinary spaces, the colimit is an interesting stack.

Since it is a colimit, there is a clear associated universal property for maps out of this stack.

It is also the stackification of the presheaf of groupoids associated to the action groupoid, hence from this description it satisfies another additional universal property for maps out of it, and there is a concrete description for maps into it. It is a fun (but tedious) exercise to see that the two universal properties are in fact the same.

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