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



Morphism of Artin stacks
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 $$\mathrm{dim}(\mathcal{F})= n+p-q\geq n-q=\mathrm{dim}(\mathcal{G}).$$


