If F is a homomorphism function, i want to ask the inverse of F is also homomorphism?

Question

In abstract algebra i have a problem to prove or disprove this statemant : "If F is a homomorphism function then the inverse of F is also homomorphism function" ?!

Answer 1

Let $f: A \to B$ be a bijective homomorphism (isomorphism) of rings (you can change the proof to other algebraic structures).

Note that you need it to be bijective to be sure that the inverse exists.

Then $f(1_A) = 1_B$, $f(a+b) = f(a)+f(b)$ and $f(ab) = f(a)f(b)$ for $a,b \in A$.

Now $$f^{-1}(1_B) = f^{-1}(f(1_A)) = 1_A.$$

Now take $c,d \in B$. Then since $f$ has an inverse (it is bijective), there exists $a,b \in A$ such that $f(a)=c$ and $f(b)=d$. In particular this gives that $a = f^{-1}(c)$ and $b = f^{-1}(d)$. It follows that

$$f^{-1}(c+d) = f^{-1}(f(a) + f(b)) = f^{-1}(f(a+b)) = a+b = f^{-1}(c) + f^{-1}(d) $$

and lastly

$$f^{-1}(cd) = f^{-1}(f(a)f(b)) = f^{-1}(f(ab)) = ab = f^{-1}(c)f^{-1}(d).$$

Answer 2

A function needs to be injective to have an inverse.

By the first isomorphism theorem if $f$ is injective, then $f$ is an isomorphism onto its image.

Because of this you can define an isomorphism as a bijective homomorphism.