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
First, in order for $\displaystyle f^{-1}$ 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 $\displaystyle f^{-1}(y)$. If f were not "one-to-one", there would be some y in B such that for both $\displaystyle x_1$ and $\displaystyle x_2$ in A, $\displaystyle f(x_1)= y= f(x_2)$. In that case we would have both $\displaystyle f^{-1}(y)= x_1$ and $\displaystyle f^{-1}(y)= x_2$, contradicting the fact that $\displaystyle f^{-1}$ is a function.
f being a homomorphism means $\displaystyle f(x_1*x_2)= f(x_1).f(x_2)$ for all $\displaystyle x_1$ and $\displaystyle x_2$ in A. To show that $\displaystyle f^{-1}$ is a homorphism, you must use that to show that $\displaystyle f^{-1}(y_1.y_2)= f^{-1}(y_1)*f^{-1}(y_2)$ for all $\displaystyle y_1$ and $\displaystyle y_2$ in B. I suggest you start by defining, for given $\displaystyle y_1$ and $\displaystyle y_2$ in B, $\displaystyle x_1= f^{-1}(y_1)$ and $\displaystyle x_2= f^{-1}(y_2)$.
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