Field Homomorphism Preserves Product Inverses

Theorem
Let $\phi: \left({F_1, +_1, \times_1}\right) \to \left({F_2, +_2, \times_2}\right)$ be a field homomorphism.

Then:
 * $\forall x \in F_1^*: \phi \left({x^{-1}}\right) = \phi \left({x}\right)^{-1}$

Proof
By definition, if $\left({F_1, +_1, \times_1}\right)$ and $\left({F_2, +_2, \times_2}\right)$ are fields then $\left({F_1^*, \times_1}\right)$ and $\left({F_2^*, \times_2}\right)$ are groups.

Again by definition:
 * the product inverse of $x$ in $\left({F_1, +_1, \times_1}\right)$ for $\times_1$ is the product inverse of $x$ in $\left({F_1^*, \times_1}\right)$
 * the product inverse of $x$ in $\left({F_2, +_2, \times_2}\right)$ for $\times_2$ is the product inverse of $x$ in $\left({F_2^*, \times_2}\right)$

The result follows from Group Homomorphism Preserves Inverses.