My question is to know whether the fibre product of $[X/G]$ by $[Y/H]$ over a base scheme is $S$ is $[X \times\_S Y/G \times H]$? And if yes, do you have any reference for it?


Thank you.



The fibre product of two quotient stacks
This is my attempt at fleshing out the details using the universal property as Scott Carnahan suggested. I think this is correct but I'm not totally sure. Let me know if theres anything off.


Let $\mathscr{X} = [X/G]$, $\mathscr{Y} = [Y/H]$ and $\mathscr{P} = [X \times\_S Y/G \times H]$. $\mathscr{X}$ is described by the following pseudofunctor:


$$
\require{AMScd}
\mathscr{X}(U) = \left\{\begin{CD} P @>>> X \\ @VVV \\ U\end{CD}\right\} 
$$


where $P \to U$ is a $G$-bundle, $P \to X$ is $G$-equivariant, and everything is compatible with the morphisms to $S$. The analogous is true for $\mathscr{Y}$ and $\mathscr{P}$ respectively. 


The fiber product $\mathscr{X} \times\_S\mathscr{Y}$ represents maps to $\mathscr{X}$ and $\mathscr{Y}$ which are equal with the compositions $S$, that is, the pseudofunctor of pairs 


$$
\left\{\begin{CD} P @>{g}>> X \\ @VVV \\ U\end{CD}, \begin{CD} Q @>{h}>> Y \\ @VVV \\ U\end{CD}\right\}
$$


such that $g$ is $G$-equivariant, $h$ is $H$-equivariant and everything is compatible over $S$. By the universal property of fiber products, this pair of diagrams is naturally equivalent to the diagram


$$
\begin{CD} P \times\_U Q @>{(g,h)}>> X \times\_S Y\\ @VVV \\ U \end{CD}.
$$


$P \times\_U Q \to U$ is a principle $G \times H$ bundle by construction and the map $(g,h)$ is a $G \times H$-equivariant morphism by construction. Therefore, the diagram is an element of $\mathscr{P}(U)$. This defines a fully faithful natural transformation of pseudofunctors $\mathscr{X} \times\_S \mathscr{Y} \to \mathscr{P}$. 


Now, it remains to show that every object in $\mathscr{P}(U)$ naturally comes from such a pair. That is, we must show that every $G \times H$ principle bundle on $U$ with a $G \times H$ equivariant morphism to $X \times\_S Y$ is the product of a $G$-bundle and an $H$-bundle with equivariant morphisms to $X$ and $Y$. 


So suppose we have a diagram


$$
\begin{CD} E @>>> X \times\_S Y\\ @VVV \\ U \end{CD} (\*)
$$


in $\mathscr{P}(U)$. $G$ and $H$ are naturally viewed as normal subgroups of $G \times H$. Then from $E$ we get a $G$-principal bundle $E/H$ and an $H$-principal bundle $E/G$. The $G \times H$ equivariant morphism $E \to X \times\_S Y$ induces a $G$ equivariant morphism $E/H \to X$ and an $H$ equivariant morphism $E/G \to Y$. These are all compatible with the morphisms to $S$ from naturality of fiber products. Therefore, we get a pair of elements


$$
\left\{\begin{CD} E/H @>>> X \\ @VVV \\ U\end{CD}, \begin{CD} E/G @>>> Y \\ @VVV \\ U\end{CD}\right\}
$$ 


in $\mathscr{X} \times\_S \mathscr{Y}(U)$. 


There is a natural morphism of principal bundles $E \to E/H \times\_U E/G$ over $U$ compatible with the maps to $X$ and $Y$ by the universal property of fiber products. Since any morphism of principal bundles is an isomorphism (equivalently since $\mathscr{P}$ is a stack so it is fibered in groupoids) then the diagram $(\*)$ is naturally isomorphic to


$$
\begin{CD} E/H \times\_U E/G @>>> X \times\_S Y\\ @VVV \\ U \end{CD}.
$$


This gives a natural equivalence of categories between $\mathscr{X} \times\_S \mathscr{Y}(U)$ and $\mathscr{P}(U)$. Therefore, we get an isomorphism of stacks $\mathscr{X} \times\_S \mathscr{Y} \cong \mathscr{P}$ over $S$. 


