Composite of Bijection with Inverse is Identity Mapping

Theorem
Let $f: S \to T$ be a bijection.

Then:
 * $f^{-1} \circ f = I_S$
 * $f \circ f^{-1} = I_T$

where $I_S$ and $I_T$ are the identity mappings on $S$ and $T$ respectively.

Proof
Let $f: S \to T$ be a bijection.

From Inverse of Bijection is Bijection, $f^{-1}$ is also a bijection.

Let $x \in S$.

From Inverse Element of Bijection:


 * $\exists y \in T: y = f \left({x}\right) \implies x = f^{-1} \left({y}\right)$

Then:

From Domain of Composite Relation and Codomain of Composite Relation, the domain and codomain of $f^{-1} \circ f$ are both $S$, and so are those of $I_S$ by definition.

So all the criteria for Equality of Mappings are met and thus $f^{-1} \circ f = I_S$.

Let $y \in T$.

From Inverse Element of Bijection:


 * $\exists x \in S: x = f^{-1} \left({y}\right) \implies y = f \left({x}\right)$

Then:

From Domain of Composite Relation and Codomain of Composite Relation, the domain and codomain of $f \circ f^{-1}$ are both $T$, and so are those of $I_T$ by definition.

So all the criteria for Equality of Mappings are met and thus $f \circ f^{-1} = I_T$.

Also see

 * Bijection iff Left and Right Inverse for the converse of this result.