# Map from schemes to stacks

I have just started studying stacks. Trying to understand the theory I was thinking about a (very interesting) toy example: $$BG$$ the classifying stack of a smooth (over a base scheme $$S$$) group G. It is well known that a (smooth) cover is given by its canonical point $$s_0: S \rightarrow BG$$ (the trivial torsor over $$S$$ seen through Yoneda). Is it true that a map from a scheme $$X$$ to $$BG$$ factor through $$s_0$$?

In particular I had in mind the concrete example where $$S=\mathrm{Spec}(\mathbb C)$$ and where $$G=GL_n$$ (so it is true that every torsor is locally trivial in the étale topology).

• When $S=\operatorname{Spec}(\mathbb{C})$ and If $X\rightarrow BG$ factors through $s_0$, then I think the corresponding family is necessarily trivial – loch Sep 24 '18 at 14:39
• @loch: If you take $X \rightarrow \mathrm{Spec}(\mathbb C)$ as the structure morphism then you're certainly right. But I was thinking about something like an endomorphism of $X$ composed with the structure morphism $x:X \rightarrow \mathrm{Spec}(\mathbb C)$ and then $s_0$. But I am not comfortable with stacks yet, so maybe I am making huge mistakes. – user192820 Sep 24 '18 at 15:47
• @loch: as stated by Sasha you were right, thank you both. – user192820 Sep 24 '18 at 15:59

## 1 Answer

No, a map $$X \to BG$$ is determined by a map $$f:X \to S$$ and a $$f^*G$$-torsor over $$X$$, while those map that factor through $$s_0$$ correspond to trivial torsors. So, as soon as there are nontrivial torsors over $$X$$, there are maps that do not factor through $$s_0$$.