# Diagonal is representable then any morphism is representable

Ariyan Javanpeykar said here in comments that,

If the diagonal is representable, then isn't any morphism $$S\rightarrow \mathcal{X}$$ with $$S$$ a scheme representable?

I could not find the statement (Thanks to my bad searching skills). I would like to prove this and use this to deduce something else.

A stack $$\mathcal{X}$$ over a scheme $$T$$ is a stack over category "schemes over $$T$$" i.e., we have a functor $$\mathcal{X}\rightarrow Sch/T$$. We can talk about $$2$$-fiber product here which I denote by $$\mathcal{X}\times_T\mathcal{X}$$.

Consider diagonal $$\Delta:\mathcal{X}\rightarrow \mathcal{X}\times_{T}\mathcal{X}$$. This is a morphism of stacks. (In stacks project, they simply write $$\mathcal{X}\rightarrow \mathcal{X}\times \mathcal{X}$$. It is somewhat confusing. May be they have fixed notation somewhere but I think specifying $$T$$ is a good idea.)

We call a morphism of stacks $$F:\mathcal{M}\rightarrow \mathcal{N}$$ to be representable if, given a morphism of stacks $$G:S\rightarrow \mathcal{N}$$, the product $$\mathcal{M}\times_{\mathcal{N}}S$$ is a scheme.

Suppose $$\Delta$$ is representable. Consider a map of stacks $$F:S\rightarrow \mathcal{X}$$. I want to see if $$F$$ is representable. For that, I take a morphism of stacks $$G:X\rightarrow \mathcal{X}$$ and prove that $$S\times_{\mathcal{X}}X$$ is a scheme.

As $$\Delta:\mathcal{X}\rightarrow \mathcal{X}\times \mathcal{X}\times_ T\mathcal{X}$$ is representable, to use the representability property of this map, I have to start with a scheme and a map from that scheme to $$\mathcal{X}\times_T \mathcal{X}$$.

I have $$F:S\rightarrow \mathcal{X}$$ and $$G:X\rightarrow \mathcal{X}$$. We can consider $$(F,G):S\times_TX \rightarrow \mathcal{X}\times_T\mathcal{X}$$. As $$S\times_TX$$ is a scheme, we can consider the map $$(F,G):S\times_TX\rightarrow \mathcal{X}\times_T\mathcal{X}$$.

As $$\Delta:\mathcal{X}\rightarrow \mathcal{X}\times_T \mathcal{X}$$ is representable, this means that $$\mathcal{X}\times_{\mathcal{X}\times_T\mathcal{X}}(S\times_TX)$$ is a scheme. I did not prove explicitly, but I am almost sure that $$\mathcal{X}\times_{\mathcal{X}\times_T\mathcal{X}}(S\times_TX)$$ is isomorphic to $$S\times_{\mathcal{X}}T$$ which is what I wanted to see.

Is this proof correct?

• Please do let me know if any statement is not clear. – Praphulla Koushik Nov 29 '18 at 11:11
• It seems like a sentence cut off in the second paragraph: "...for any object $C$ of $\mathcal{C}$..." – WSL Nov 29 '18 at 11:20
• @WSL : Does it look better now.? Thanks for pointing out.. – Praphulla Koushik Nov 29 '18 at 11:26

I guess it is correct (and may be rendered in a simpler way). Ideed, let $$\delta:\mathcal{X}\rightarrow \mathcal{X}\times_{T}\mathcal{X}$$ be the diagonal map. If it is representable then every morphism $$u : S → \mathcal{X}$$ is representable. For $$v : V → \mathcal{X}$$ another morphism with $$V$$ a scheme, we have that $$S \times_{\mathcal{X}} V \cong \mathcal{X} \times_{\mathcal{X}\times\mathcal{X}} (S \times_T V)$$ is 1-isomorphic to a scheme ($$\delta$$ is representable) and this 1-isomorphism turns out to be an isomorphism because $$S \times_{\mathcal{X}} V$$ is in fact a category fibered in sets, therefore corresponds to a scheme.