# Map from a classifying space to a stack

Let $$G$$ be an algebraic group over a field $$k$$, and let $$BG$$ be its classifying space. Let $$X$$ be a stack over $$k$$ (e.g. $$X$$ could be the Picard stack $$Pic(S)$$, for some scheme $$S$$). I'm trying to understand what it means to have a morphism $$BG \to X$$ of stacks. Since $$BG=[pt/G]$$, my guess is that a map $$BG \to X$$ is just an object $$x \in X(k)$$ together with a homomorphism of groups $$G(k) \to Aut(x)$$. Is this correct? This doesn't seem to take into account any of the geometric structure of $$G$$, so I wouldn't be surprised if this is isn't right.

What is the right way to think of maps out of a classifying stack?

You're almost there! The problem is that, as you've surmised, the group $$\mathrm{Aut}(x)$$ does not capture enough of the geometric structure of $$G$$. But that's easily solved:

For every $$x\in X$$ we can define a sheaf of groups $$\underline{\mathrm{Aut}}(x)$$ such that for every $$k$$-scheme $$S$$ $$\underline{\mathrm{Aut}}(x)(S):=\mathrm{Aut}_{X(S)}(x_S)\,.$$ where $$x_S$$ is the image of $$x$$ under the functor $$X(k)→X(S)$$ induced by the map of schemes $$S\to \mathrm{Spec}\,k$$. That this is a sheaf follows from the definition of stack ("descent for morphisms"), the details might depend on your preferred way of phrasing the definition.

Then, for every sheaf of groups $$G$$ we can define a stack $$BG$$, such that for every $$k$$-scheme $$S$$ $$BG(S)$$ is the groupoid of $$G_S$$-torsors over $$S$$ (this recovers the notion when $$G$$ is a group scheme). Note that $$BG(k)$$ has a distinguished object $$\ast$$ (corresponding to the trivial $$G$$-torsor over $$k$$) such that there's a canonical isomorphism of sheaves of groups $$\underline{\mathrm{Aut}}(\ast)\cong G\,.$$

Then the groupoid of maps $$BG\to X$$ is the groupoid of pairs $$(x\in X(k),\varphi:G\to \underline{\mathrm{Aut}}(x))$$, where $$x$$ is an object of $$X(k)$$ and $$\varphi$$ is a map of sheaves of groups. The morphisms in the groupoid are the morphisms in $$X(k)$$ that conjugate the maps from $$G$$.

The map in one direction is easy: it sends $$F:BG\to X$$ to the pair $$(F(\ast),G\cong \underline{\mathrm{Aut}}(\ast)\to \underline{\mathrm{Aut}}(F(\ast)))$$.

The map in the other direction is slightly trickier: suppose I have a pair $$(x,\varphi)$$. Then for every $$S$$ I have to give a functor $$BG(S)\to X(S)$$. Essentially this sends a $$G_S$$-torsor $$T$$ to $$T\times_{G_S} x_S$$, where $$G_S$$ acts on $$x_S$$ via $$\varphi$$. Making this precise is a bit annoying and hence I shall leave it as an exercise.

Really, the easy way of conceptualizing this (and arguably proving it) is to notice that $$BG$$ is the stackification of the prestack $$BG^{pre}$$ sending $$S$$ to the groupoid with one objects and $$G(S)$$ automorphisms. Then we can compute the groupoid of maps $$BG\to X$$ as the groupoid of maps $$BG^{pre}\to X$$, and the latter is easily seen to coincide with my description above.

For the homotopy theorists that might be listening: yes the above could be simply summarized by saying that the functor from groups to pointed groupoids sending $$G$$ to $$BG$$ is the left adjoint to the functor sending $$X$$ to $$\Omega X$$.

All I described here works in an arbitrary topos: I did not use that the sheaf of groups $$G$$ came from a group scheme.