Group Homomorphism Maps Inverses to Inverses

We would like to show that given a homomorphism $f: G \rightarrow H$, where $G$ and $H$ are groups, the identity maps to the identity. Namely, $f(e_{G}) = e_{H}$, where $e_{G} \in G$ and $e_{H} \in H$.

$$f(g e_{G}) = f(g)f(e_{G})$$ $$\implies f(g) = f(g) f(e_{G})$$ $$\implies f(g)^{-1} f(g) = f(g)^{-1} f(g) f(e_{G})$$ $$\implies e_{H} = f(e_{G})$$

Let $f$ be a homomorphism from $G \rightarrow H$, where $G$ and $H$ are groups. We want to show: $f(g^{-1}) = f^{-1}(g)$.

$$f(gg^{-1}) = f(g)f(g^{-1})$$ $$\implies f(e_{G}) = f(g)f(g^{-1})$$ $$f(g)^{-1} f(e_{G}) = f(g)^{-1} f(g)f(g^{-1})$$ $$\implies f(g)^{-1} f(e_{G}) = f(g^{-1})$$

from the previous proof, we know $f(e_{G}) = e_{H}$. Namely, under $f$, the identity maps to the identity. Hence, we can say:

$$f(g)^{-1} e_{H}= f(g^{-1})$$ $$\implies f(g)^{-1} = f(g^{-1})$$