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factorization system
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.
When there is a factorization system $(\mathcal{E},\mathcal{M})$ on a category $\mathcal{C}$, we say that $\mathcal{C}$ has $(\mathcal{E},\mathcal{M})$factorization, and an $(\mathcal{E},\mathcal{M})$factorization of a morphism $f$ is a factorization of $f$: $f=m\circ e$, such that $m\in\mathcal{M}$ and $e\in\mathcal{E}$.
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
$\xymatrix@+=1.5cm{A\ar[dr]^{f}\ar[r]^{s}\ar[d]_{t}&C\ar[d]^{u}\\ D\ar[r]_{v}&B}$ 
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
$\xymatrix@+=1.5cm{A\ar[r]^{s}\ar[d]_{t}&C\ar[d]^{u}\ar@<0.5ex>[dl]_{g}\\ D\ar[r]_{v}\ar@<0.5ex>[ur]_{h}&B}$ 
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
$\xymatrix@+=1.5cm{A\ar[r]^{s}\ar[d]_{s}&C\ar[d]^{u}\ar[dl]{h\circ g}\\ C\ar[r]_{u}&B}$ 
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…
References
 1 F. Borceux Basic Category Theory, Handbook of Categorical Algebra I, Cambridge University Press, Cambridge (1994)
Mathematics Subject Classification
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