Recall that any function $f:A\to B$ can be factored as $h\circ g$ where $g:A\to f(A)$ is a surjection and $h:f(A)\to B$ is an injection. This phenomenon is true in many mathematical systems: homomorphisms between groups, rings, lattices, continuous maps between topological spaces, etc...

However, in the setting of category theory, while it is still true that a morphism can be factored into the composition of two morphisms (one of them, say, being the identity morphism), the fact that one factor is an epimorphism and the other a monomorphism no longer holds in general. Categories where such kinds of factorizations exist is of great interest. Factorization of morphisms in a category can be formalized as follows:

Definition. Let $\mathcal{C}$ be a category. An ordered pair $(\mathcal{E},\mathcal{M})$ of classes of morphisms in $\mathcal{C}$ is called a factorization system if

• every morphism $f$ in $\mathcal{C}$ can be ``factored'' as $f=m\circ e$ where $m\in \mathcal{M}$ and $e\in \mathcal{E}$ ,
• $\mathcal{E}$ is orthogonal to $\mathcal{M}$ : $\mathcal{E} \perp \mathcal{M}$ ,
• every isomorphism is in both $\mathcal{E}$ and $\mathcal{M}$ , and
• both $\mathcal{E}$ and $\mathcal{M}$ are closed under composition; in other words, if $x,y$ are both in one class and $x\circ y$ is defined, then $x\circ y$ is in that class too.

One of the first properties of having a factorization system $(\mathcal{E},\mathcal{M})$ is that the $(\mathcal{E},\mathcal{M})$ -factorization of a morphism $f$ is unique up to isomorphism:

Proposition 1   If we have a commutative diagram

where $s,t\in \mathcal{E}$ and $u,v\in \mathcal{M}$ , then $C\cong D$ .

Proof. Because $\mathcal{E}\perp \mathcal{M}$ , there are unique morphisms $g:C\to D$ and $h:D\to C$ such that the diagram

is commutative. Then $h\circ g\circ s = h \circ t = s$ and $u \circ h\circ g = v \circ g = u$ , which means we have a commutative diagram

But $s\perp u$ , so the morphism $h\circ g: C\to C$ making the above diagram commute is uniquely determined. Since $1_C:C \to C$ is another morphism making the diagram commute, we must have $h\circ g=1_C$ . Similarly, one sees that $g\circ h=1_D$ . This implies that $C\cong D$ .

More to come...

Bibliography

1

F. Borceux Basic Category Theory, Handbook of Categorical Algebra I, Cambridge University Press, Cambridge (1994)