# Abstract Algebra/Group Theory/Homomorphism/Homomorphism Maps Inverse to Inverse

 0.   $f({\color{Blue}g}) \circledast f({\color{BrickRed}g^{-1}}) = f({\color{Blue}g} \ast {\color{BrickRed}g^{-1}})$ f is a homomorphism 1.   $f({\color{Blue}g}) \circledast f({\color{BrickRed}g^{-1}}) = f({\color{Blue}e_{G}})$ definition of inverse in G . 2.   $f({\color{Blue}g}) \circledast f({\color{BrickRed}g^{-1}}) = {\color{OliveGreen}e_{K}}$ homomorphism f maps identity to identity 3.   ${\color{BrickRed} [} f({\color{Blue}g}) {\color{BrickRed}]^{-1}} \circledast f({\color{Blue}g}) \circledast f({\color{BrickRed}g^{-1}}) = {\color{BrickRed} [} f({\color{Blue}g}) {\color{BrickRed}]^{-1}} \circledast {\color{OliveGreen}e_{K}}$ as f(g) is in K, so is its inverse [f(g)]-1 . 4.   ${\color{OliveGreen}e_{K}} \circledast f({\color{BrickRed}g^{-1}}) = {\color{BrickRed} [} f({\color{Blue}g}) {\color{BrickRed}]^{-1}}$ inverse on K, eK is identity of K 5.   $f({\color{BrickRed}g^{-1}}) = {\color{BrickRed} [} f({\color{Blue}g}) {\color{BrickRed}]^{-1}}$ eK is identity of K