# Thread: If f is a homomorphism, is it true that f-1(x+y)=f-1(x)+f-1(y)?

1. ## If f is a homomorphism, is it true that f-1(x+y)=f-1(x)+f-1(y)?

If f is a homomorphism, is it true that f-1(x+y)=f-1(x)+f-1(y) and f-1(xy) = f-1(x) * f-1(y)??

2. ## Re: If f is a homomorphism, is it true that f-1(x+y)=f-1(x)+f-1(y)?

Originally Posted by gummy_ratz
If f is a homomorphism, is it true that f-1(x+y)=f-1(x)+f-1(y) and f-1(xy) = f-1(x) * f-1(y)??
Let me see if I can understand. You have a bijective ring homomorphism $f:R\to S$, and you are asking if $f^{-1}$ is a ring homomorphism? The answer is yes since $f(f^{-1}(x)+f^{-1}(y))=f(f^{-1}(x))+f(f^{-1}=x+y$ and so $f^{-1}(x+y)=f^{-1}(x)+f^{-1}(y)$, the multiplicative part follows similarly.

3. ## Re: If f is a homomorphism, is it true that f-1(x+y)=f-1(x)+f-1(y)?

What if we don't know the homomorphism is bijective? The question just says, f:R->S is a ring homomorphism.

4. ## Re: If f is a homomorphism, is it true that f-1(x+y)=f-1(x)+f-1(y)?

If a function is not bijective then the inverse is not defined.(There are things that will fail if a function is not bijective. for example having a left inverse is related to being one-to-one, having a right inverse is related to being surjective, if a function isn't a bijection then it won't have inverses(it can be proved that the inverse of a function is unique easily)) Hence, it doesn't make sense to talk about the inverse of a homomorphism if it's not defined. in fact homomorphisms that are bijective are special, they are called isomorphisms as you know.

5. ## Re: If f is a homomorphism, is it true that f-1(x+y)=f-1(x)+f-1(y)?

Originally Posted by gummy_ratz
If f is a homomorphism, is it true that f-1(x+y)=f-1(x)+f-1(y) and f-1(xy) = f-1(x) * f-1(y)??
If f is a ring homomorphism with kernel K, then the condition $f^{-1}(x+y)=f^{-1}(x)+f^{-1}(y)$ is equivalent to $K = K+K$, and the condition $f^{-1}(xy) = f^{-1}(x) * f^{-1}(y)$ implies that $K = K*K.$ The first of those is true, the second one is not.

For example, if R is a ring, S is the ring R[x] of polynomials over R, and the homomorphism f:S→R is the map $p(x)\mapsto p(0)$, then K is the ideal of polynomials with zero constant term. But $K*K$ consists of those elements of K for which the coefficient of x is also 0. Therefore $f^{-1}(0*0)\supsetneqq f^{-1}(0)*f^{-1}(0).$