# On Wikipedia's definition of zero morphisms

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 $$\bar 0_{XZ}$$.

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 $$f,g$$.
• 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. Commented Apr 29 at 13:21