non-Hopfian groups  Let $A\to B$ and $B\to C$ be epimorphisms of Abelian (infinitely generated) groups. Let $A\cong C$. Is then $A\cong B$?
 A: Replacing $C$ by $A$, you ask the following: if there are epis $A \twoheadrightarrow B$ and $B \twoheadrightarrow A$, is there an iso $A \cong B$ ? The answer is no, take $A=\mathbb{Z}^{\oplus \mathbb{N}}$ and $B=A \times \mathbb{Q}$. There is an obvious projection $B \to A$. On the other hand, since $\mathbb{Q}$ is countable, there is an epimorphism $\mathbb{Z}^{\oplus 2\mathbb{N}} \to \mathbb{Q}$, which may be multiplied with an isomorphism $\mathbb{Z}^{\oplus 2\mathbb{N}+1} \cong \mathbb{Z}^{\oplus \mathbb{N}}$ to optain an epimorphism $A \to B$. But $A$ and $B$ are not isomorphic since $d(A)=0$ and $d(B)=\mathbb{Q}$, where $d(-)$ denotes the largest divisible subgroup.
For results in the positive direction, see the paper "Correct classes of modules" by Robert Wisbauer, online. Especially the notion of epi-correct classes is what you might be looking for.
A: Take $A$ be the free Abelian group of countable rank, i.e. the infinite direct product (or as Andreas Blass prefers to call it, the direct sum) of infinite cyclic groups, $B$ be the direct product (sum) of $\mathbb{Q}$ and $A$, $C=A$. Since $A$ is free Abelian, there exists a surjective homomorphism 
$A\to B$. There is also a projection $B\to C$ but $A$ is obviously not isomorphic to $B$. 
