Exercise 2.3.5 (Homomorphic inverse)

Solution: Since $K$ is a field, and the fact that $0\ne a\in K$, we have that $a$ is a unit, $a{a}^{-1}=1$, and $a^{-1}\in K$. By Lemma 2.3.3, we have that $\varphi :K\to L$ is injective. Hence, $\varphi (a)\varphi (a^{-1})=\varphi (a\circ a^{-1})=\varphi (1)=1$, and $\varphi (a^{-1})\varphi (a)=\varphi (a^{-1}\circ a)=\varphi (1)=1$. Therefore we have that $\varphi (a^{-1})$ is both a left and right inverse of $a$ and hence it is the only inverse of $a$. Therefore, by injectivity $\varphi (a^{-1})=\varphi {(a)}^{-1}$.