For any Group homomorphism f:G→H, we have f(g−1)=f(g)−1.
Indeed, f(g−1)f(g)=f(g−1g)=f(eG)=eH, and similarly for multiplication on the left.
https://arbital.com/p/group_homomorphism_image_of_inverse
by Patrick Stevens Jun 14 2016 updated Jun 14 2016
The operations of "taking inverses" and "applying a group homomorphism" commute: it does not matter in which order we do them.
For any Group homomorphism f:G→H, we have f(g−1)=f(g)−1.
Indeed, f(g−1)f(g)=f(g−1g)=f(eG)=eH, and similarly for multiplication on the left.