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Thread: projective, module homomorphism

  1. #1
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    projective, module homomorphism

    A module P is projective if for every surjective module homomorphism $\displaystyle f : N \twoheadrightarrow M$ and every module homomorphism $\displaystyle g : P \rightarrow M$, there exists a homomorphism $\displaystyle h : P \rightarrow N$ such that $\displaystyle f \circ h = g$.

    Prove that if $\displaystyle P$ is projective and $\displaystyle h: P \rightarrow M$ is a surjective $\displaystyle R-$module homomorphism, then there exists some $\displaystyle R-$module homomorphism $\displaystyle s: M \rightarrow P$ such that $\displaystyle h \circ s=1_M$.

    This is a section map. I don't know how to prove this. I was thinking to let $\displaystyle N=M$, but it doesn't seem that easy.
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  2. #2
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    Quote Originally Posted by zelda2139 View Post

    Prove that if $\displaystyle P$ is projective and $\displaystyle h: P \rightarrow M$ is a surjective $\displaystyle R-$module homomorphism, then there exists some $\displaystyle R-$module homomorphism $\displaystyle s: M \rightarrow P$ such that $\displaystyle h \circ s=1_M$.
    the question, as you wrote it, is wrong! (check what you wrote carefully!) the correct one is this:

    Prove that if $\displaystyle P$ is projective and $\displaystyle h: M \rightarrow P$ is a surjective $\displaystyle R-$module homomorphism, then there exists some $\displaystyle R-$module homomorphism $\displaystyle s: P \rightarrow M$ such that $\displaystyle h \circ s=1_P$.

    the proof is trivial: we have the identity map $\displaystyle 1_p: P \longrightarrow P$ and a surjection $\displaystyle h: P \longrightarrow M.$ thus by the definition of projectivity there exists a map $\displaystyle s: P \longrightarrow M$ such that $\displaystyle hs=1_P.$
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