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?
Best Answer
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]