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Problem 6
Recall that if $f:A \rightarrow B$ is a one-to-one function mapping $A$ onto $B$, then $f^{-1}(b)$ is the unique $a \in A$ such that $f(a)=b$. Prove that if $\phi: S \rightarrow S'$ is an isomorphism of $\langle S, *\rangle$ with $\langle S', *' \rangle$, then $\phi^{-1}$ is an isomorphism of $\langle S', *' \rangle$ with $\langle S, * \rangle$.
Solution
One-to-One: Let $a',b',c'\in S'$ and suppose $\phi^{-1}(a')=\phi^{-1}(b')$. Now $a'=\phi(\phi^{-1}(a'))=\phi(\phi^{-1}(b'))=b'$. So $\phi^{-1}$ is one-to-one.
Onto: Let $a\in S$. Then $\phi^{-1}(\phi(a))=a$. So $\phi^{-1}$ maps $S'$ onto $S$.
Homomorphism: Let $a',b'\in S$. Then $\phi(\phi^{-1}(a'*'b'))=a'*'b'$. Since $\phi$ is an isomorphism, $\phi(\phi^{-1}(a')*\phi^{-1}(b'))=\phi(\phi^{-1}(a'))*'\phi(\phi^{-1}(b'))=a'*'b'$ and because it is one-to-one, $\phi^{-1}(a'*b')=\phi^{-1}(a')*'\phi^{-1}(b')$.