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.