The fibre product of two quotient stacks 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.
 A: 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 $S$. 
