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Dimension of preimage question!

  1. Mar 17, 2009 #1
    let g be the inverse function of f (not necessarily a bijective inverse function)
    If S is any subset of W then the pre-image of S is

    g(S) = {v ε V: f(v) ε S}

    Suppose that U is a subspace of W.
    Prove that g(U) is a subspace of V.
    Also prove that
    dim( g(U) )= dim( U intersect Im(f) ) + dim( ker (f) )

    For the first part,
    by linearity, f(av1+v2)=af(v1)+f(v2)=au1+ u2 ε U where v1ε g(U), v2ε g(U)
    so av1+v2 ε g(U)---> a g(u1)+g(u2)ε g(U)
    if f(v)=v.then g(U) is not empty
    therefore, g(U) is a subspace.

    For the second part,
    I am not sure. can i just simply assume that U=W? then g(U) = V and U intersect im(f) is im(f) so clearly dim(V)=dim imf + dim kerf

    Could anyone verify my first proof and give me a hint for the second part? Thanks in advance.
  2. jcsd
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