Composite of Bijection with Inverse is Identity Mapping

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Theorem

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

Then:

\(\ds f^{-1} \circ f\) \(=\) \(\ds I_S\)
\(\ds f \circ f^{-1}\) \(=\) \(\ds I_T\)

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:

$\exists y \in T: y = \map f x \implies x = \map {f^{-1} } y$


Then:

\(\ds \map {f^{-1} \circ f} x\) \(=\) \(\ds \map {f^{-1} } {\map f x}\) Definition of Composition of Mappings
\(\ds \) \(=\) \(\ds \map {f^{-1} } y\) by hypothesis
\(\ds \) \(=\) \(\ds x\) by hypothesis
\(\ds \) \(=\) \(\ds \map {I_S} x\) Definition of Identity Mapping


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:

$\exists x \in S: x = \map {f^{-1} } y \implies y = \map f x$


Then:

\(\ds \map {f \circ f^{-1} } y\) \(=\) \(\ds \map f {\map {f^{-1} } y}\) Definition of Composition of Mappings
\(\ds \) \(=\) \(\ds \map f x\) by hypothesis
\(\ds \) \(=\) \(\ds y\) by hypothesis
\(\ds \) \(=\) \(\ds \map {I_T} y\) Definition of Identity Mapping


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 $f \circ f^{-1} = I_T$.

$\blacksquare$


Also see


Sources

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