Composite of Mapping with Inverse
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Theorem
Let $f: S \to T$ be a mapping.
Then:
- $\forall x \in S: \map {f^{-1} \circ f} x = \eqclass x {\RR_f}$
where:
- $\RR_f$ is the equivalence induced by $f$
- $\eqclass x {\RR_f}$ is the $\RR_f$-equivalence class of $x$.
Proof
Let $y = \map f x$.
Then by the definition of induced equivalence:
- $x \in \eqclass x {\RR_f}$
By the definition of the inverse of a mapping:
- $f^{-1} = \set {\tuple {y, x}: \tuple {x, y} \in f}$
Thus:
- $\eqclass x {\RR_f} = \set {s \in \Dom f: \map f s = \map f x}$
By definition:
- $\map {f^{-1} } y = \eqclass x {\RR_f}$
Hence the result.
$\blacksquare$
Also see
Sources
- 1955: John L. Kelley: General Topology ... (previous) ... (next): Chapter $0$: Functions
- 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{I}$: Sets and Functions: Problem $\text{AA}$: Relations