Inverse Mapping in Induced Structure

Theorem

Let $\left({S, \circ}\right)$ be an algebraic structure.

Let $\left({T, \oplus}\right)$ be an abelian group.

Let $f$ be a homomorphism from $S$ into $T$.

Let $f^*$ be the induced structure inverse of $f$.

Then $f^*$ is a homomorphism from $\left({S, \circ}\right)$ into $\left({T, \oplus}\right)$.

Proof

Let $\left({T, \oplus}\right)$ be an abelian group.

Let $x, y \in S$.

Then:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle f^* \left({x \circ y}\right)\)	\(=\)	\(\displaystyle \left({f \left({x \circ y}\right)}\right)^{-1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Induced Structure Inverse
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({f \left({x}\right) \oplus f \left({y}\right)}\right)^{-1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	as $f$ is a homomorphism
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({f \left({y}\right) \oplus f \left({x}\right)}\right)^{-1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Commutativity of $\oplus$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({f \left({x}\right)}\right)^{-1} \oplus \left({f \left({y}\right)}\right)^{-1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Inverse of a Group Product
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle f^* \left({x}\right) \oplus f^* \left({y}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Induced Structure Inverse

$\blacksquare$

Sources

• Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 13$: Theorem $13.7: \ 2^\circ$