Inverse Morphism is Unique
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Theorem
Let $\mathbf C$ be a metacategory.
Let $f: C \to D$ be an isomorphism of $\mathbf C$.
Then $f$ admits a unique inverse morphism $g: D \to C$.
Proof
Since $f$ is an isomorphism, it admits at least one inverse morphism.
Now let $g, g': D \to C$ be two inverse morphisms for $f$.
Then:
\(\ds g\) | \(=\) | \(\ds g \circ \operatorname{id}_D\) | Axiom $(C2)$ for metacategories | |||||||||||
\(\ds \) | \(=\) | \(\ds g \circ \left({f \circ g'}\right)\) | $g'$ is an inverse morphism for $f$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \left({g \circ f}\right) \circ g'\) | Axiom $(C3)$ for metacategories | |||||||||||
\(\ds \) | \(=\) | \(\ds {\operatorname{id}_C} \circ g'\) | $g$ is an inverse morphism for $f$ | |||||||||||
\(\ds \) | \(=\) | \(\ds g'\) | Axiom $(C2)$ for metacategories |
In conclusion, $g = g'$.
Hence the result.
$\blacksquare$
Sources
- 2010: Steve Awodey: Category Theory (2nd ed.) ... (previous): $\S 2.8$: Exercise $2.3$