# 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:

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

$\blacksquare$