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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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up vote 3 down vote accepted

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

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