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

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

    Quote Originally Posted by gummy_ratz View Post
    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.
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    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.
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    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.
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    Re: If f is a homomorphism, is it true that f-1(x+y)=f-1(x)+f-1(y)?

    Quote Originally Posted by gummy_ratz View Post
    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).
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