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