Zero Morphism does not Depend on Zero Object

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Theorem

Let $\mathbf C$ be a category.

Let $A$ and $B$ be objects of $\mathbf C$.

Let $0_1$ and $0_2$ be zero objects of $\mathbf C$.


Then the morphism defined as the composition

$\beta \circ \alpha : A \to 0_1 \to B$

of the unique morphism $\alpha : A \to 0_1$ and the unique morphism $\beta : 0_1 \to B$ is equal to the morphism defined as the composition

$\delta \circ \gamma : A \to 0_2 \to B$

of the unique morphism $\gamma : A \to 0_2$ and the unique morphism $\delta : 0_2 \to B$.


Proof

There are unique morphisms $\epsilon : 0_1 \to 0_2$ and $\zeta : 0_2 \to 0_1$.

Since $0_1$ is terminal, we have

$\zeta \circ \epsilon = \operatorname{id}_{0_1}$
$\beta \circ \zeta = \delta$

Since $0_2$ is terminal, we have

$\epsilon \circ \alpha = \gamma$

Hence

\(\ds \beta \circ \alpha\) \(=\) \(\ds \beta \circ \operatorname{id}_{0_1} \circ \alpha\)
\(\ds \) \(=\) \(\ds \beta \circ \zeta \circ \epsilon \circ \alpha\)
\(\ds \) \(=\) \(\ds \delta \circ \gamma\)

$\blacksquare$

Notation

This justifies the notation $0 := \beta \circ \alpha$, whenever a zero object exists in $\mathbf C$.