First of all, thank you to everyone for helping!
So, I am currently familiarizing myself with category theory, in particular with abelian categories. In MacLane's Categories for the Working Mathematician he gives the following characterisation of an exact sequence in an abelian category.
"A sequence $A \overset{f}{\rightarrow} B \overset{g}{\rightarrow}C$ is exact at $B$ if and only if $gf = 0$ and to every $z \in_m C$ there exists a $y \in_m B$ with $gy \equiv z$" (Theorem 3 (v), p. 205)
where $x \in_m X$ denotes a member of $X$, i.e. an equivalence class of the equivalence relation $\equiv$ induced by the partial ordering $\leqq$, where we have that $x \leqq y$ for two morphism with codomain $X$ if and only if there exists a morphism $z$ such that $x$ factors through $y$, i.e., $x = yz$.
When he proves that the condition on the members together wit $gf = 0$ implies that the sequence is exact at $B$, he uses the mono-epic factorisation $f = me$ and states
"Conversely, given this property of all $y \in_m B$, take $k = \text{ker }g$; then $k\in_m B$ and $gk = 0$. Therefore there is a member $x \in_m A$ with $fx \equiv k$; that is, with $ku = mexv$ for suitable epis $u$ and $v$."
So far, everything is clear for me, but then he goes on
"But this equation implies that the monic $k$ factors through $m$, and hence that $\text{im }f \geqq \text{ker }g$."
And here comes my question: why can he deduce that $k$ factors through $m$ because so far we only know that $ku$ factors through $m$? How can he get rid of the $u$ on the left-hand side?
Best Answer
Consider the epi-mono factorization $ex = nw$ (where $n$ is mono and $w$ is epi). Then the equation $ku = mexv = (mn)(wv)$ gives two epi-mono factorizations of a single morphism, so we must have $k\cong mn$ as objects over $b$, i.e., $k$ factors through $m$.