Given a representable, surjective morphism of Artin stacks $\phi:\mathcal{F}\to\mathcal{G}$, is it true (like it happens for schemes) that $\dim\mathcal{G}\leq\dim\mathcal{F}$?

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 $nq$ 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 $$\mathrm{dim}(\mathcal{F})= n+pq\geq nq=\mathrm{dim}(\mathcal{G}).$$ 


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? 

