Inverse of a homomorphic function

**mathrld** · August 17th 2012, 06:25 AM

Let f(A,∗)-->(B,⋅) be homomorphic. Prove that if f−1 is a function then it is homomorphic
Inorder for f−1 to be a function then f must be a bijection

**HallsofIvy** · August 17th 2012, 06:51 AM

First, in order for $f^{-1}$ to exist, f must be "onto". That is, for every y in B there must be x in A such that f(x)= y. Otherwise we could not define $f^{-1}(y)$ . If f were not "one-to-one", there would be some y in B such that for both $x_1$ and $x_2$ in A, $f(x_1)= y= f(x_2)$ . In that case we would have both $f^{-1}(y)= x_1$ and $f^{-1}(y)= x_2$ , contradicting the fact that $f^{-1}$ is a function.

f being a homomorphism means $f(x_1*x_2)= f(x_1).f(x_2)$ for all $x_1$ and $x_2$ in A. To show that $f^{-1}$ is a homorphism, you must use that to show that $f^{-1}(y_1.y_2)= f^{-1}(y_1)*f^{-1}(y_2)$ for all $y_1$ and $y_2$ in B. I suggest you start by defining, for given $y_1$ and $y_2$ in B, $x_1= f^{-1}(y_1)$ and $x_2= f^{-1}(y_2)$ .