Non-isomorphic algebraic structures such that each surjects homomorphically onto the other Off the top of my head, I cannot think of any algebraic structures $X$ and $Y$ such that each surjects homomorphically onto the other, yet $X$ and $Y$ are non-isomorphic. What are some examples of this kind of thing?
 A: Take two abelian groups $X=\mathbb{Z}/2\mathbb{Z}\times\prod\limits_{i=0}^\infty \mathbb{Z}$ and $Y=\mathbb{Z}/3\mathbb{Z}\times\prod\limits_{i=0}^\infty \mathbb{Z}$. I think they satisfy the required property.
The projections are the simplest ones, with the torsion component goes to $0$, the first $\mathbb{Z}$-component projects onto the first torsion component.
A: According to Tom Leinster's answer to this question on Math Overflow, it's a result of A. L. S. Corner that there exists an Abelian group $A$ isomorphic to $A^3$ but not to $A^2$. Of course this means that $A$ and $A^2$ are homomorphic images of each other without being isomorphic.
A: Find two compact Hausdorff spaces $X, Y$ which embed into each other but which are not homeomorphic, for example $[0, 1]$ and $[0, 1] \cup [2, 3]$. Embeddings $X \to Y$ of compact Hausdorff spaces induce surjections $C(Y) \to C(X)$ on C*-algebras of continuous complex-valued functions (by Tietze extension) and by Gelfand-Naimark you get a counterexample, a priori in C*-algebras but in fact in $\mathbb{C}$-algebras (using the fact that the norm on a C*-algebra is determined by its algebra structure). 
