Is monomorphism going in both directions sufficient for isomorphism? In category theory, it seems that a monomorphism from $A$ to $B$ and one from $B$ to $A$ should be enough to guarantee isomorphy, but it doesn't seem to be so. (If I'm right then there's something fishy with the standard definition of "subobject")
So here's the counterexample I thought up, please explain where I went wrong.
Consider a category consisting of 2 objects $A$ and $B$. There is a monomorphism $\phi: A \to B$ and another $\psi : B \to A$. "Close" this under composition in much the same way you do when defining a free group (that is, no non-trivial identities are allowed). I claim that this does not guarantee isomorphism. All morphisms are monic, since no identities hold, so the condition for monomorphism is trivially satisfied.
What am I doing wrong here?
 A: Dear Seamus, an example of non-isomorphic objects mutually monomorphing into each other is the following, in the category of groups ( I haven't tried to follow your sketch of construction).
Consider the free group on two generators $F_2$. Its commutator subgroup $C\subset F_2$ is a free group on denumerably many generators: $C=F_\infty$. This can be proved elegantly by using topological covering spaces [you can look it up in Massey's Introduction to Algebraic Topology for example]. 
So you have monomorphisms 
$F_2 \hookrightarrow F_\infty$ and $F_\infty \hookrightarrow  F_2$, although  $F_2$ and  $F_\infty$ are not isomorphic, since their abelianizations are free $\mathbb Z$ modules on respectively two and denumerably many generators. 
I have used that monomorphisms in the category of groups coincide with injective morphisms, which is a not trivial but true result [ Jacobson, Basic Algebra, vol.II, Prop 1.1]
A: Your counterexample is correct; indeed it is the universal one, every other counterexample comes from a functor defined on your category. For a counterexample in the category of fields, see my answer here Counterexamples in Algebra?.
You seem to be worried about subobjects. If $X$ is an object and $U,V \leq X$ are subobjects such that $U \leq V$ and $V \leq U$, then $U = V$. The reason is that the morphisms $U \to V$ and $V \to U$ over $X$ are uniquely determined (since $V \to X, U \to X$ are monomorphisms). Likewise are the compositions $U \to V \to U, V \to U \to V$ uniquely determined, namely the identity. Thus $U = V$. Thus you don't get into trouble.
A: I should state first that this reply has only to do with the above mentioned ideas in the category of models of a first order theory. 
John Goodrick's work is referenced in Joel's post above, and I have heard John Goodrick speak about this at least once. Specifically, John mentioned the following (and a lot more that I didn't write down):
Fix some countable, complete first-order theory, $T$. Suppose $T$ has the following property: Whenever we are given two models $\mathcal M_1$ and $\mathcal M_2$ of $T$ which have elementary embeddings into each other, then  $\mathcal M_1 \cong \mathcal M_2.$
Then $T$ is superstable and nonmultidimensional (and I know if John replies to this, he can mention many other things, but I don't remember now). In the case that $T$ is actually $\omega -$stable, nonmultidimensional implies the bi-embedding property stated in the above paragraph. 
A: One of the simplest example , lightest in structure and very easy in checking is : 
In the category of monoids take the canonical injection $i$ from $(N,+,0)$ to $(Z,+,0)$.
This is a monomorphism that is also an epimorphism yet not an iso ($i$ is not a surjection).
($N$ and $Z$ are the positive integers and integers respectively)
