Inverse Morphism is Unique

From ProofWiki
Jump to: navigation, search


Let $\mathbf C$ be a metacategory.

Let $f: C \to D$ be an isomorphism of $\mathbf C$.

Then $f$ admits a unique inverse morphism $g: D \to C$.


Since $f$ is an isomorphism, it admits at least one inverse morphism.

Now let $g, g': D \to C$ be two inverse morphisms for $f$.


\(\displaystyle g\) \(=\) \(\displaystyle g \circ \operatorname{id}_D\) $\quad$ Axiom $(C2)$ for metacategories $\quad$
\(\displaystyle \) \(=\) \(\displaystyle g \circ \left({f \circ g'}\right)\) $\quad$ $g'$ is an inverse morphism for $f$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left({g \circ f}\right) \circ g'\) $\quad$ Axiom $(C3)$ for metacategories $\quad$
\(\displaystyle \) \(=\) \(\displaystyle {\operatorname{id}_C} \circ g'\) $\quad$ $g$ is an inverse morphism for $f$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle g'\) $\quad$ Axiom $(C2)$ for metacategories $\quad$

In conclusion, $g = g'$.

Hence the result.