Let $f:A \mapsto B$, $g:B \mapsto A$ and $h:B \mapsto B$ be such that $g \circ f=\operatorname{id}_A$ and $f \circ g \circ h=\operatorname{id}_B=h \circ f \circ g$. Can we conclude $h=\operatorname{id}_B$? Are there any "naturally occurring" categories where this is false?
We certainly have that $f$ and $g$ are bimorphisms (i.e. simultaneously monomorphisms and epimorphisms), and that $h$ is an automorphism of $B$ (with $h^{-1}=f \circ g$).
In a balanced category such as $\mathbf{Set}$, every bimorphism would be an isomorphism, hence $f$ would have an inverse $f^{-1}$, and we could conclude $f^{-1}=g$ from $g \circ f=\operatorname{id}_A$, which would show $h=\operatorname{id}_B$.
The simplest way to construct a category where $h\neq\operatorname{id}_B$ seems to be to take $\operatorname{hom}(A,A)=\{\operatorname{id}_A\}$, $\operatorname{hom}(B,B)=\{\operatorname{id}_B,h\}$, $\operatorname{hom}(B,A)=\{g\}$ and $\operatorname{hom}(A,B)=\{f,h \circ f\}$. So there are categories where $h=\operatorname{id}_B$ is false. What about a counterexample for the case $A=B$? Edit: Constructing a counterexample without first generating the set of all "valid" relations between $f$, $g$ and $h$ is difficult, because consequences/violations of associativity can easily be overlooked. But if we know the set of all "valid" relations, then we already know the answer.
I have the impression that none of the standard examples of non-balanced categories like posets, quivers, category of rings, or topological categories could provide an example where $h=\operatorname{id}_B$ is false. Is there a name for such an "almost" isomorphism between $A$ and $B$ as provided by $f$, $g$ and $h$, as well as a name for the categories where such "almost" isomorphism are always a real isomorphism?
Notice that $gh=1gh=gfgh=g1=g$, so that $fg=fgh=1$. It follows that $f$ and $g$ are inverse isomorphisms.
The relation $hfg=1$ has not been used here, so it can be omitted.