Composite of Inverse of Mapping with Mapping
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Theorem
Let $f: S \to T$ be a mapping.
Then:
- $f \circ f^{-1} = I_{\Img f}$
where:
- $f \circ f^{-1}$ is the composite of $f$ and $f^{-1}$
- $f^{-1}$ is the inverse of $f$
- $I_{\Img f}$ 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:
- $f \circ f^{-1} := \set {\tuple {x, z} \in T \times T: \exists y \in S: \tuple {x, y} \in f^{-1} \land \tuple {y, z} \in f}$
Thus $f \circ f^{-1}$ is a relation on $T \times T$.
Let $x \in \Img f$.
Then:
- $\exists y \in S: \tuple {x, y} \in f^{-1}$
As $f$ is a mapping, and so by definition a many-to-one relation;
- $\map f y = x$
for all $y$ such that $\tuple {x, y} \in f^{-1}$.
That is:
- $\forall x \in \Img f: f \circ \map {f^{-1} } x = x$
Hence the result by definition of identity mapping.
$\blacksquare$
Also see
Sources
- 1955: John L. Kelley: General Topology ... (previous) ... (next): Chapter $0$: Functions