Image of Element under Inverse Mapping
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Theorem
Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping such that its inverse $f^{-1}: T \to S$ is also a mapping.
Then:
- $\forall x \in S, y \in T: \map f x = y \iff \map {f^{-1} } y = x$
Proof
Sufficient Condition
Let $f: S \to T$ be a mapping.
From the definition of inverse mapping:
- $f^{-1} = \set {\tuple {y, x}: \tuple {x, y} \in f}$
Let $y = f \paren x$.
From the definition of the preimage of an element:
- $\map {f^{-1} } y = \set {s \in S: \tuple {y, x} \in f}$
Thus:
- $x \in \map {f^{-1} } y$
However, $f^{-1}$ is a mapping.
Therefore, by definition:
- $\forall y \in T: \tuple {y, x_1} \in f^{-1} \land \tuple {y, x_2} \in f^{-1} \implies x_1 = x_2$
Thus:
- $\forall s \in f^{-1} \paren y: s = x$
Thus:
- $\map {f^{-1} } y = \set x$
That is:
- $x = \map {f^{-1} } y$
$\Box$
Necessary Condition
Let $\map {f^{-1} } y = x$.
By definition of inverse mapping:
- $\map f x = y$
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 5$: Composites and Inverses of Functions: Theorem $5.2$
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 22.3$: Injections; bijections; inverse of a bijection