Krdytkyu

Ariyan Javanpeykar said here in comments that,

If the diagonal is representable, then isn't any morphism $Srightarrow mathcal{X}$ 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 $mathcal{X}times_Tmathcal{X}$.

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 $G:Srightarrow mathcal{N}$, the product $mathcal{M}times_{mathcal{N}}S$ is a scheme.

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

As $Delta:mathcal{X}rightarrow mathcal{X}times mathcal{X}times_ Tmathcal{X}$ 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 $F:Srightarrow mathcal{X}$ and $G:Xrightarrow mathcal{X}$. We can consider $(F,G):Stimes_TXrightarrow mathcal{X}_Tmathcal{X}$. As $Stimes_TX$ is a scheme, we can consider the map $(F,G):Stimes_TXrightarrow mathcal{X}times_Tmathcal{X}$.

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

Is this proof correct?

Any comments are welcome.

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
$$
S times_{mathcal{X}} V cong mathcal{X} times_{mathcal{X}timesmathcal{X}} (S times_T V)
$$
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.