When does one invertibility condition suffices? It is often the case that, to prove $f$ being the inverse morphism of $g$, one has only to show $fg = id$ and the other direction ($gf = id$) is guaranteed to be true -- e. g. when considering vector space morphisms.
In other words, "f is right inverse of g or left inverse" implies "f is right and left inverse of g". I will name it "one-suffices property".
Can we say something about this property in a category-theoretical context? For example a statement "Ever category fulfilling property ... fulfills the one-suffices property too".
PS: I thought that I had asked this question before, but none of my questions seem to have this topic.
 A: For a fixed $n \in \mathbb{N}$, the category of $n$-dimensional vector spaces has this property, but not the category of all finite-dimensional vector spaces (consider $K \hookrightarrow K^2 \twoheadrightarrow K$), let alone the category of all vector spaces. So this does not happen so often as you might think. It happens more often in one-object categories, which are more commonly known as monoids. Also, the $\mathbf{Ab}$-enriched one-object categories are more commonly known as rings. Trivially, every commutative monoid and ring has this property. But also every left Artinian ring or monoid (in particular, every finite ring) has this property: If $fg=1$ in $R$, then in the sequence of left ideals $R \supseteq Rf \supseteq Rf^2 \supseteq \dotsc$ we have $R f^n = R f^{n+1}$ for some $n$, and multiplication with $g^n$ on the right yields $R = R f$, and now $gf=1$ follows easily.
A: It's worth noting that your example for vector spaces is missing some conditions. Namely, the domain and codomain vector spaces have to be finite-dimensional and have the same dimension. Otherwise, you get counterexamples like the "shift right" and "shift left" maps from $\mathbb{R}^\mathbb{N}$ to itself. For finite dimensional spaces, consider the matrices
$$
A = \begin{pmatrix}
1 & 0 & 0\\
0 & 1 & 0
\end{pmatrix}
$$
and
$$
B = \begin{pmatrix}
1 & 0\\
0 & 1\\
0 & 0
\end{pmatrix}
$$
Then $AB = I$, but $BA \neq I$.
So a refinement might be: if $f \colon A \to B$ and $g \colon B \to A$ are morphisms and there exists an isomorphism between $A$ and $B$, when does $fg = \mathrm{id}_B$ imply that $gf = \mathrm{id}_A$.
We already saw than in the case of vector spaces, a sufficient extra ingredient is finite dimensionality. A similar condition works for sets: if $A$ and $B$ are finite sets, this works.
More generally, if the category is equipped with a faithful functor to the category of sets, a sufficient condition is that the images of $A$ and $B$ (under the faithful functor) be finite sets. This follows from the corresponding statement for finite sets and faithful functorality. This then applies to a large number of examples, such as groups, topological spaces, etc. with their underlying set functor.
Note, however, that this is merely a sufficient condition, not a necessary condition. Indeed, vector spaces have an underlying set, but unless the ground field is finite, a finite dimensional vector space rarely has a finite underlying set.
For finite dimensional vector spaces, the key idea is that if $fg = \mathrm{id}_B$, then $g$ is injective and so $B$ isomorphic to $g$'s image in $A$. But for a finite dimensional vector space, a subspace that's isomorphic to the whole space must be equal to the whole space. Thus, $g$ is also surjective. For vector spaces, $g$ being both injective and surjective implies that $g$ is an isomorphism. In general, any right inverse is equal to any left inverse, so since $f$ is a left inverse of $g$, it's equal to $g$'s right inverse (which exists because $g$ is an isomorphism).
So what's the categorical version of all that? Just to simplify things, we'll consider $A$ = $B$ and for the general version, you'll just need to pre- and post-compose things with the given isomorphism.
$fg = \mathrm{id}_A$ means that $g$ is a split monomorphism. Split monomorphisms generally speaking the strongest kind of monomorphism: any split monomorphism is (whenever it makes sense) normal, regular and strong.
This is all to say that $g$ embeds $A$ into $A$ in the nicest possible way. It's a normal subgroup in the category of groups, it's a subspace inclusion in the category of topological spaces, etc.
But then we can repeat this process and embed $A$ as a subobject of $A$ as a subobject of $A$. We have a whole chain $\mathrm{id}_A \supseteq g \supseteq g^2 \supseteq g^3 \supseteq g^4 \ldots$ of subobjects of $A$.
By definition, if $A$ is artinian, then this chain stabilizes, which forces $g$ to be an isomorphism: if $g^{n + 1} \hookrightarrow g^n$ is an isomorphism for some n, then since this inclusion is just $g$, $g$ is an isomorphism. So if $g$ has an inverse $h$, then $f = f(gh) = (fg)h = h$, so $f$ is also a right inverse for g.
So what kinds of objects are artinian in this sense? Finite dimensional vector spaces are all artinian. Finite sets are artinian. For one confusing case, a ring is called artinian if its module over itself is artinian in the above sense. I'm not sure if there's a name for a ring satisfying a descending chain condition on its subrings (which would be an artinian object in the category of rings).
But remember that our chain of subobjects was actually a chain of split monomorphisms, and those are nicer than arbitrary subobjects. For example, in the category of groups, they'd be normal subgroup inclusions. In particular, simple groups satisfy the property we want, since they only have trivial normal subgroups. Any descending chain of normal subgroups of a simple group is either always the whole group or is eventually always the trivial subgroup.
I'm sure there are other natural classes of objects that satisfy this "descending chain condition of split subobjects".
