Composite of Inverse of Mapping with Mapping

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

Then:
 * $f \circ f^{-1} = I_{\operatorname {Im} \left({f}\right)}$

where:
 * $f \circ f^{-1}$ is the composite of $f$ and $f^{-1}$
 * $f^{-1}$ is the inverse of $f$
 * $I_{\operatorname {Im} \left({f}\right)}$ is the identity mapping on the image set of $f$.

Proof
By Inverse of Mapping is One-to-Many Relation, $f^{-1}$ is a one-to-many relation:
 * $f^{-1} \subseteq T \times S$

whose domain is the image set of $f$.

By definition of composition of relations:


 * $f \circ f^{-1} := \left\{{\left({x, z}\right) \in T \times T: \exists y \in S: \left({x, y}\right) \in \mathcal f^{-1} \land \left({y, z}\right) \in f}\right\}$

Thus $f \circ f^{-1}$ is a relation on $T \times T$.

Let $x \in \operatorname {Im} \left({f}\right)$.

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

As $f$ is a mapping, and so by definition a many-to-one relation;
 * $f \left({y}\right) = x$

for all $y$ such that $\left({x, y}\right) \in f^{-1}$.

That is:
 * $\forall x \in \operatorname {Im} \left({f}\right): f \circ f^{-1} \left({x}\right) = x$

Hence the result by definition of identity mapping.