Inverse Mapping in Induced Structure

From ProofWiki
Jump to navigation Jump to search

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

$\blacksquare$


Sources