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 $\mathbf C$.