Diagonal is representable then any morphism is representable

Question

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?

Any comments are welcome.

Answer 1

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.

For your second question, I don't see any special use of the category of schemes, the displayed isomorphism is a basic categorical fact.