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Let ${\bf C}$ be a category with zero morphisms, i.e., for each $X,Y\in {\bf C}$, there is a morphism $0_{XY}:X\rightarrow Y$ satisfying certain properties ($0_{XY}$ composes with every morphism from $Y$ or to $X$ to give another such). According to Wikipedia, "the collection of $0_{XY}$ is unique." Does it mean that given $X,Y\in{\bf C}$, there can be only one morphism that we can use as $0_{XY}$? When defining kernel and cokernel, one uses these morphisms $0_{XY}$, so it appears as though kernel and cokernel depends on a particular choice of the family of $0_{XY}$, which is very unpleasant.

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    $\begingroup$ A example of citogensis in the wild. $\endgroup$
    – PyRulez
    Jan 17, 2015 at 1:31

3 Answers 3

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Yes. The idea is the same as the proof of the uniqueness of identities in a monoid. If $f_{X, Y} : X \to Y$ is a family of zero morphisms and $g_{X, Y} : X \to Y$ is another family of zero morphisms, then

$$f_{Y, Z} \circ g_{X, Y} = g_{X, Z} = f_{X, Z}$$

for every triple of objects $X, Y, Z$.

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  • $\begingroup$ @almagest: no, I don't use that anywhere. If you're worried about the use of $Y$, you can set $Y = X$ or $Y = Z$. $\endgroup$ Sep 18, 2019 at 18:59
  • $\begingroup$ @almagest: I don't know what you mean by "a single different zero morphism" here. The definition of what it means for $f_{A, B}$ to be a zero morphism requires zero morphisms $f_{X, A}$ and $f_{B, X}$ for every other object $X$. $\endgroup$ Sep 19, 2019 at 6:22
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Zero morphism in a category $\mathcal C$ with Zero object $O$ from an object $A$ to $B$ which factors through zero object,that is the following diagram hold:

$A\rightarrow O \to B$

Note that this factorization is unique as $O$ is both intial and terminal object.So zero morphism from an object $A$ to $B$ is unique.

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    $\begingroup$ Yes, it is (almost) obvious that a zero object gives you a complete set of zero morphisms which compose nicely. But the OP's definition (and Wikipedia's) does not define the set of zero morphisms that way. So for your answer to work, you would need to show that a zero morphism exists. Note also that a category may have multiple zero objects (albeit isomorphic). $\endgroup$
    – almagest
    Sep 18, 2019 at 18:55
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Q: Does it mean that given $X,Y\in{\bf C}$, there can be only one morphism that we can use as $0_{XY}$?

Yes. Provided that the $0_{XY}$ form a complete collection, ie (i) given any (possibly equal) objects $X,Y$ in $\mathbf{C}$ there is a zero arrow $0_{XY}$, and (ii) these compose, ie $0_{YZ}\circ 0_{XY}=0_{XZ}$, then there are no other zero arrows.

Suppose that such a complete collection exists, and that for some objects $A,B$ the arrow $f_{AB}$ is a zero arrow. We have $0_{AB}=0_{BB}\circ 0_{AB}=0_{BB}\circ f_{AB}$ using (i) and (ii). But then $0_{BB}\circ f_{AB}=\text{Id}_B\circ f_{AB}=f_{AB}$ since $f_{AB}$ is a zero. Hence $f_{AB}=0_{AB}$. $\Box$

Note that one way of getting zero arrows is to find a zero object. But: (A) a category may have multiple zero objects. But different zero objects give the same zero arrows; and (B) we may have no zero objects, but still have a complete set of zero arrows.

[Trivial example: one object and two arrows, the identity and the zero arrow. Less trivial: two objects, a complete set of zero arrows (4 of them), distinct identities (2 of them), and one other arrow (source one object, target the other).]

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