Morphism of Artin stacks Given a representable, surjective morphism of Artin stacks $\phi:\mathcal{F}\to\mathcal{G}$, is it true (like it happens for schemes) that $\dim\mathcal{G}\leq\dim\mathcal{F}$?
 A: No.  $B\mathbb{G}_m\to\mathrm{pt}$ is surjective with source of dimension -1.
As commented below this is nonsense, since this is not representable. 
Let me try to make amends for my flip response to the question.  Of course what I am about to write could be equally stupid.  However, $\phi$ surjective means that the induced map $\mathcal{F}\times_{\mathcal{G}}G\to G$ is surjective, where $G\to \mathcal{G}$ is smooth surjective with $G$ a scheme. Suppose that the dimension of $G$ is $n$ and that of the scheme $\mathcal{F}\times_{\mathcal{G}}G$ is $n+p$ where $p\geq 0$. Then the dimension of $\mathcal{G}$ is $n-q$ where $q$ is the relative dimension of the smooth surjective morphism $G\times_{\mathcal{G}}G\to G$.  Now $\mathcal{F}\times_{\mathcal{G}}G\to\mathcal{F}$ is smooth and surjective.  Furthermore:
$$(\mathcal{F}\times_{\mathcal{G}}G)\times_\mathcal{F}(\mathcal{F}\times_{\mathcal{G}}G)\simeq (\mathcal{F}\times_{\mathcal{G}}G)\times_G(G\times_\mathcal{G}G)$$
and hence is of relative dimension $q$ over $(\mathcal{F}\times_{\mathcal{G}}G)$.  Therefore $$\mathrm{dim}(\mathcal{F})= n+p-q\geq n-q=\mathrm{dim}(\mathcal{G}).$$
A: This is not a complete answer, but should hopefully be a start.  Given a point $x \in \mathcal{F}$, the dimension of $\mathcal{F}$ at $x$ is given by picking an atlas $X$ of $\mathcal{F}$, and then computing
\begin{equation}
dim_x(\mathcal{F}) = dim_x(X) - dim(Aut_{\mathcal{F}}(x)).
\end{equation}
The dimension of $\mathcal{F}$ is then the supremum of its dimension at all its points.  Given surjective, representable $\phi:\mathcal{F} \to \mathcal{G}$, we obtain a surjective map of atlases $X \to Y$.  Therefore $dim_x(X) \geq dim_{\phi(x)}(Y)$ for all points $x$.
So, we're reduced to comparing $Aut_{\mathcal{F}}(x)$ to $Aut_{\mathcal{G}}(\phi(x))$.  There is necessarily a surjection $Aut_{\mathcal{F}}(x) \to Aut_{\mathcal{G}}(\phi(x))$, so there is a question of how much larger the automorphisms of $x$ in $\mathcal{F}$ are compared to those of $\phi(x)$ in $\mathcal{G}$.  That is, we would need to verify that
\begin{equation}
dim_x(X) - dim_{\phi(x)}(Y) \geq dim(Aut_{\mathcal{F}}(x)) - dim(Aut_{\mathcal{G}}(\phi(x)))
\end{equation}
holds for all points $x$.  I feel like this should be true just because my possibly faulty intuition says that the relative dimension of the automorphisms of a surjective, representable map won't exceed the relative dimension of the atlases.  However, I don't see how to prove it right now.
Does anyone know how to prove this, or provide a counterexample?
