Q:

Image factorisations in abelian categories

Let $\mathcal{C}$ be an abelian category and $f:A\longrightarrow B$ a morphism of $\mathcal{C}$. Suppose you have a factorisation $f=i\circ p$, with $p$ a strong epi, $i$ a mono and $\operatorname{cod}(p)=\operatorname{dom}(i)$ "the" cokernel of the kernel of $f$.
hen by duality uou get a second factorisation $f=i'\circ p'$ with $p'$ a strong epi, $i'$ a mono and $\operatorname{cod}(p')=\operatorname{dom}(i')$ the kernel of the cokernel of $f$.
Could you please explain the meaning of the second paragraph, the one starting with "by duality" till the end?
Second question: call $I=\operatorname{cod}(p)=\operatorname{dom}(i)$ and $I'=\operatorname{cod}(p')=\operatorname{dom}(i')$. Can I say $I=I'$ or only that $I$ and $I'$ are isomorphic?

A:

The duality will become meaningful only if we assume the factorization for all morphisms $f$ in all Abelian categories $\mathcal C$. 
So, it's a theorem, and since being Abelian is a self-dual condition ($\mathcal C^{op}$ will be Abelian again), the dual of the theorem will hold (that is the theorem applied to $\mathcal C^{op}$).
No, only isomorphic: $I\cong I'$.

