For $A \leq B$ groups, there exists an epimorphism from $B$ to $A$. I think this is false because of this counterexample:
For $A_3 \leq S_3$, the only homomorphism $\varphi: S_3 \mapsto A_3$ is the zero one, because every transposition has to get mapped to $id$ and they generate the whole image of $S_3$.
Are there some other easy counter examples?
 A: First of all you can check that due to the fundamental theorem of finitely generated abelian groups you can always construct such epimorphism if $B$ is finitely generated and abelian.
So you have to search for examples elsewhere. The smallest one possible is $S_3$. Note that coincidentally it is also the smallest non-abelian group. In that sense there is no better example then the one you've provided.
The other way is to look at infinitely generated groups. In this scenario you can even go with abelian. For example there is no epimorphism $\mathbb{Q}\to\mathbb{Z}$ (see this) because $\mathbb{Q}$ is divisible while $\mathbb{Z}$ is not. IMO this is by far the easiest to visualize/imagine cause we are so used to integers and rationals.
Another crazy example is if you take a finitely generated group and an infinitely generated subgroup. Yes, such pathologies exist, but not in the abelian world. Take $\mathbb{F}(x,y)$ a free group on 2 generators and define
$$H_k=<x^nyx^{-n}\ |\ n\geq k>$$
It can be shown that $H_k$ is not finitely generated and as such it cannot be a homomorphic image of $\mathbb{F}(x,y)$.
