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I think the answer is yes because of the following immediate (and easy) facts:

  1. The morphisms associated with initial (terminal) objects are right (left) zeroes.
  2. If $r$ is a right zero and $l$ a left zero, then $lr$, if defined, is a zero morphism.
  3. If $z$ is a zero morphism, then $gzf$, whenever defined, is also a zero morphism.

Now, if we have an initial object $I$ and a terminal object $T$, we can define $0_{X, Y} := X\to I\to T\to I\to Y$ which thereby form a compatible family of zero morphisms.

Is my reasoning correct?

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In your definition of $0_{X,Y}$, you compose maps $X\to I$ and $T\to I$. Why should these maps exist?

For example, in the category of sets, there is no function from the terminal object $1$ (a one-point set) to the initial object $\varnothing$ (the empty set), so this category does not have zero morphisms: there is no possible choice for the zero morphism $0_{1,\varnothing}$.

In fact, if $C$ is a category with an initial object $0$ and a terminal object $1$, then $C$ has zero morphisms if and only if $1\cong 0$.

In one direction, if $1\cong 0$, then there is an arrow $z_{1,0}\colon 1\to 0$, and then for any objects $X$ and $Y$, we can define $z_{X,Y} = {{!}\circ z_{1,0}\circ {!}}\colon X\to 1\to 0\to Y$ and check that this gives a compatible family of zero morphisms.

Conversely, if $C$ has zero morphisms, then in particular there is a (zero) morphism $1\to 0$, but any arrow from a terminal object to an initial object is an isomorphism (e.g., see here).

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  • $\begingroup$ Oof, how embarrassing! My bad. $\endgroup$
    – Atom
    Dec 10, 2023 at 5:21

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