# 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?

1. 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 $\mathcal C^{op}$).
2. No, only isomorphic: $I\cong I'$.
• Thanks for answering. I totally got the point 1. For the 2., if I have a theorem in $\mathcal{C}$, then I go to the opposite category, the object $I$ shouldn't remain the same in $\mathcal{C}^{op}$? – bateman Feb 28 '17 at 21:46
• It's not the same object. $I$ comes from the image factorization in $C, I'$, from that in the dual of $C$. – Kevin Carlson Feb 28 '17 at 23:14