This entry deals with basic properties of monomorphisms and related notions (as extremal and regular monomorphisms, retractions etc) as well as the dual notions.

Monomorphisms (epimorphisms, bimorphisms) are closed under composition.

Proof. a) $g\circ f\circ h=g\circ f\circ k$ $f\circ h=f\circ k$ $h=k$
b) $h\circ g\circ f=k\circ g\circ f$ $h\circ g=k\circ g$ $h=k$
c) An easy corollary of a) and b).

Proof. $g\circ h=g\circ k$ $f\circ g\circ h=f\circ g\circ k$ $h=k$

Retractions and sections are closed under composition.

Proof. Suppose we are given $\Map hBA$ , $\Map kCB$ such that $f\circ h=id_B$ , $g\circ k=id_C$ . Then $(g\circ f)\circ(h\circ k)=g\circ(f\circ h)\circ k=g\circ id_B\circ k=g\circ k=id_C$ . Thus we have shown the first part of the claim. The second part is dual to the first one and the third one follows from the first two.

Proof. If $g\circ f$ is a section then there exists a morphism $h$ such that $h\circ(g\circ f)=(h\circ g)\circ f=id_A$ , thus $f$ is a section as well.

Proof. Let $\Map fAB$ be a section and $\Map gBA$ be the left inverse to $f$ , i.e., $\Map gBA$ , $g\circ f=id_A$ . If $f\circ h=f\circ k$ then $h=id_A\circ h=g\circ f\circ h=g\circ f\circ k=id_A\circ k=k$ . The duality principle yields the second part of the claim.

Recall that a morphism is called an isomorphism if it is a section and a retraction at the same time.

Proof. $h=id_A\circ h=(g\circ f)\circ h=g\circ(f\circ h)=g\circ id_B=g$

Proof. The implication $\boxed{\Leftarrow}$ is obvious. The implication follows from the above lemma.

The morphism $g$ from the above proposition is called the inverse of $f$ and denoted by $f^{-1}$ .

As an easy corollary we get: