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Wikipedia defines $f : X \to Y$ to be a zero morphism if $(1)$ $gf = hf$ for any object $Z$ and $g, h:Y \to Z$, and $(2)$ $fg = fh$ for any object $W$ and any $g, h : W \to X$. It then defines a category having zero morphisms to be a category with a fixed morphism $0_{XY} : X \to Y$ for each pair of objects $X$ and $Y$ such that a certain diagram commutes, in particular giving $f \circ 0_{XY} = 0_{XZ} = 0_{YZ} \circ g$ for all objects $X$, $Y$, and $Z$ and morphisms $f : Y\to Z$ and $g : X \to Y$.

The Wikipedia article claims, and I have verified, that the $0_{--}$ morphisms are in fact zero morphisms (as a consequence of the defining diagram) and are unique as a set of morphisms making the diagram commute. But then it claims

This way of defining a "zero morphism" and the phrase "a category with zero morphisms" separately is unfortunate, but if each hom-set has a "zero morphism", then the category "has zero morphisms".

Is this correct? In other words, if there is a zero morphism $\bar0_{XY} : X \to Y$ for every pair of objects $X$ and $Y$, then must these morphisms make the preceding diagram commute? I was able to show that the "outer edges" of the diagram commute, namely $f \circ \bar 0_{XY} = \bar 0_{YZ} \circ g$. But I am not sure whether these equal $\bar 0_{XZ}$.

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Note first that zero morphisms are unique: if $0, 0' : X \to Y$ are both zero morphisms then $$ 0 = 1_Y\circ 0 = 0'\circ h\circ 0 = 0'\circ 1_X = 0' $$ so long as there exists $h : Y \to X$. The second equality is (1) and the third equality is (2).

(If there are no arrows $Y \to X$ then I am unsure whether or not $0 = 0'$.)

Note that we have arrows in both directions between any two objects; in particular, because $\bar 0_{ZX}$ is an arrow $Z \to X$, the zero morphisms are unique and it suffices to show that $z := f\circ\bar0_{XY} = \bar0_{YZ}\circ g$ is a zero morphism whence it is necessarily equal to $\bar0_{XZ}$. So we prove each property in turn:

  1. Let $h, h' : Z \to W$. Then $$ h\circ z = h\circ f\circ\bar0_{XY} = h'\circ f\circ\bar0_{XY} = h'\circ z $$ because $\bar0_{XY}$ is a zero morphism.
  2. Let $h, h' : W \to X$. Then $$ z\circ h = \bar0_{YZ}\circ g\circ h = \bar0_{YZ}\circ g\circ h' = z\circ h' $$ because $\bar0_{YZ}$ is a zero morphism.

Thus $z$ is a zero morphism and so $z = \bar0_{XZ}$ and $$ f\circ\bar0_{XY} = \bar0_{XZ} = \bar0_{YZ}\circ g $$ for all $f,g$.

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  • $\begingroup$ Very nice. In fact, if there are no arrows $Y \to X$, it can happen that zero morphisms are non-unique. Zhen Lin gives a simple example here. $\endgroup$
    – WillG
    Commented Apr 29 at 13:21

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