Translating a diagram chase into an element-free proof One part of the four lemma says that:

Consider the following diagram with exact rows in an abelian category $\mathsf{A}$:

If $m$ and $p$ are monomorphisms and $l$ is an epimorphism, then $n$ is a monomorphism.

We can very simply prove this by diagram chasing. I'll copy a proof below:

*

*Let $c \in C$ be such that $n(c) = 0$.

*$t(n(c))$ is then $0$.

*By commutativity, $p(h(c)) = 0$.

*Since $p$ is injective, $h(c) = 0$.

*By exactness, there is an element $b$ of $B$ such that $g(b) = c$.

*By commutativity, $s(m(b)) = n(g(b)) = n(c) = 0$.

*By exactness, there is then an element $a′$ of $A′$ such that $r(a′) = m(b)$.

*Since $l$ is surjective, there is $a$ in $A$ such that $l(a) = a′$.

*By commutativity, $m(f(a)) = r(l(a)) = m(b)$.

*Since $m$ is injective, $f(a) = b$.

*So $c = g(f(a))$.

*Since the composition of $g$ and $f$ is trivial, $c = 0$.

I want to "translate" this proof into an element-free proof using universal properties.
The beginning is simple. The composition $\ker n\to C\to C'\to D'$ is zero. By commutativity of the diagram, so is $\ker n\to C\to D\to D'$. But $p:D\to D'$ is monic, so we conclude that $\ker n\to C\to D$ is zero. This seems to be the first 4 steps in the proof above.
However I don't know how to continue. (If an element-free proof of this cannot simply be a translation of the diagram chase, I would still want to learn it.)

Thanks to the answer and the comment by Jackozee Hakkiuz, I did the following proof. (The notations are not exactly the same, but it should be clear. Also the proposition 1.3.3 is simply the fact that the pullback of an epimorphism is epic.)



I hope that this may be useful to someone :)
