Inversion Mapping is Mapping
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Theorem
Let $\struct {G, \circ}$ be a group.
Let $\iota: G \to G$ be the inversion mapping on $G$.
Then $\iota$ is indeed a mapping.
Proof
To show that $\iota$ is a mapping, it is sufficient to show that:
- $\map \iota a \ne \map \iota b \implies a \ne b$:
\(\ds \map \iota a\) | \(\ne\) | \(\ds \map \iota b\) | by hypothesis | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds a^{-1}\) | \(\ne\) | \(\ds b^{-1}\) | Definition of $\iota$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds a \circ a^{-1} \circ b\) | \(\ne\) | \(\ds a \circ b^{-1} \circ b\) | Cancellation Laws and Group Axiom $\text G 1$: Associativity | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds e \circ b\) | \(\ne\) | \(\ds a \circ e\) | Group Axiom $\text G 3$: Existence of Inverse Element | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds b\) | \(\ne\) | \(\ds a\) | Group Axiom $\text G 2$: Existence of Identity Element |
$\blacksquare$