Composite of Bijection with Inverse is Identity Mapping

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

Then:


 * $$f^{-1} \circ f = I_S$$, where $$I_S$$ is the identity mapping on $$S$$;


 * $$f \circ f^{-1} = I_T$$, where $$I_T$$ is the identity mapping on $$T$$.

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

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


 * Let $$x \in S, y = f \left({x}\right) \implies x = f^{-1} \left({y}\right)$$ (from Inverse Element of Bijection).

Then:

$$ $$ $$ $$

From Domain of Composite Relation and Range of Composite Relation, the domain and range 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 $$y \in T, x = f^{-1} \left({y}\right) \implies y = f \left({x}\right)$$ (from Inverse Element of Bijection).

Then:

$$ $$ $$ $$

From Domain of Composite Relation and Range of Composite Relation, the domain and range 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 $$f \circ f^{-1} = I_T$$.