Injective Morphisms, Monomorphisms and Left Invertible Morphisms in Abelian Categories

Let $\mathcal{C}$ be an abelian category. A morphism $f:X \rightarrow Y$ is called injective if its kernel is zero. $f$ is called monomorphism if whenever $f \circ g=0$, for $g:Z \rightarrow X$, then $g=0$. We have the result that a morphism is injective if and only if it is a monomorphism. My question is: what is the correct terminology for the stronger property of existence of a morphism $h: Y \rightarrow X$ such that $h \circ f=id_{X}$? What is the minimal additional assumption that we need to make for $\mathcal{C}$ such that a morphism is injective if and only if the above-mentioned property is true? For example, if the objects are sets, then the equivalence is true.

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$f$ is then said left invertible. – Berci Nov 10 '12 at 23:10
Your final claim is false for trivial reasons: the inclusion $\emptyset \to X$ is always injective/monic but splits if and only if $X$ is also empty. – Zhen Lin Nov 11 '12 at 0:01

1 Answer

$f$ is left invertible. Equivalently, it is a split monomorphism. I wrote a blog post on the subject that you might find helpful.

The condition that every monomorphism splits is quite strong. For example, in $\text{Top}$, the split monomorphisms are precisely the inclusions of retracts, and most monomorphisms in $\text{Top}$ are not of this form. Similarly, in $\text{Ab}$, the split monomorphisms are precisely the inclusions of direct summands, and most monomorphisms in $\text{Ab}$ are not of this form.

For abelian categories, $\mathcal{A}$ has this property if and only if every short exact sequence splits, hence if and only if $\mathcal{A}$ is semisimple. This is a highly restrictive condition.

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