# Factoring a morphism via its Co-image and Image in Abelian Categories

This question refers to the Lemma 3.10 of the chapter of Homological Algebra of the Stacks Project. In particular, the lemma states that any morphism $f:x \rightarrow y$ can be factored uniquely as $x\rightarrow Coim(f) \rightarrow Im(f) \rightarrow y$. Where i am having a difficulty, is understanding the argument that gives the morphism $Coim(f) \rightarrow Im(f)$. The proof of the lemma, says "the morphism $Coim(f) \rightarrow Y \rightarrow Coker(f)$ must be zero because it is the unique morphism that gives rise to the zero morphism $x\rightarrow y \rightarrow Coker(f)$. Where does this uniqueness follow from?

Alternatively, i can see that we should have that the morphism $ker(f) \rightarrow x \rightarrow Im(f)$ should be zero, but i still can not give an argument.

Finally, i can see that in the category of modules the above situation is immediate.

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Let's distinguish: for objects ${\rm Ker},{\rm Im},{\rm Coker},{\rm Coim}$ I write capital letters, for the corrsponding morphisms small letters. So that ${\rm ker}f:{\rm Ker}f\to x$ is the equalizer of $f$ and $0$, and ${\rm coker}f:y\to{\rm Coker}f$ is the coequalizer of $f$ and $0$.

Then, by definition, ${\rm im}f={\rm ker}({\rm coker}f)$, so that equalizes ${\rm coker}f$ and $0$, i.e. ${\rm im} f\cdot {\rm coker}f=0$ (writing composition from left to right), and in particular, as $f\cdot {\rm coker}f= 0$, there is a unique $f_1:x\to{\rm Im}f$ such that $f_1\cdot {\rm im}f=f$.

Now, ${\rm coim}f={\rm coker}({\rm ker}f)$, and $0={\rm ker} f\cdot f = {\rm ker}f\cdot f_1\cdot{\rm im}f$. Now use that ${\rm im}f$ is monomorphism, we get that ${\rm ker}f\cdot f_1=0$ as wished, and thus there is a unique $f_2$ such that $f_1={\rm coim}f\cdot f_2$, so briefly, this 'uniqueness' you asked tracks back to the definition of (co-)equalizer.

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Thanks for the nice answer. What i was missing was that $imf$ is a monomorphism. – Manos Nov 10 '12 at 21:11