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Exact Sequences - Lifting Homomorphisms - D&F Ch 10 - Theorem 28

  1. May 12, 2014 #1
    I am reading Dummit and Foote, Chapter 10, Section 10.5, Exact Sequences - Projective, Injective and Flat Modules.

    I need help with a minor step of D&F, Chapter 10, Theorem 28 on liftings of homomorphisms.

    In the proof of the first part of the theorem (see image below) D&F make the following statement:

    -------------------------------------------------------------------------------

    Conversely, if F is in the image of [itex] \psi'[/itex] then [itex] F = \psi' (F') [/itex] for some [itex] F' \in Hom_R (D, L) [/itex] and so [itex] \phi ( F (d) )) = \phi ( \psi ( F' (d))) [/itex] for any [itex]d \in D[/itex]. ...

    ----------------------------------------------------------------------------

    My problem is that surely [itex] F = \psi' (F') [/itex] implies that [itex] \phi ( F (d) )) = \phi ( \psi' ( F' (d))) [/itex] and NOT [itex] \phi ( F (d) )) = \phi ( \psi ( F' (d))) [/itex]?

    Hoping someone can help>

    Theorem 28 and the first part of the proof read as follows:

    attachment.php?attachmentid=69697&stc=1&d=1399873986.png

    Peter
     

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    Last edited: May 12, 2014
  2. jcsd
  3. May 12, 2014 #2

    micromass

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    This cannot be true. We know that ##F^\prime\in \textrm{Hom}(D,L)##, and thus ##F^\prime(d)\in L##. But the domain of ##\psi^\prime## is not ##L##, but rather ##\textrm{Hom}(D,L)##. So ##\psi^\prime(F^\prime(d))## makes no sense since ##F^\prime(d)## is not in the domain of ##\psi^\prime##.
     
  4. May 12, 2014 #3
    Thanks ...

    Yes, you are right ...

    BUT, now I have two problems ...

    Problem 1 ... to get [itex] \phi ( F (d) )) = \phi ( \psi' ( F' (d))) [/itex] I simply replaced the [itex]F[/itex] with [itex] \psi' (F') [/itex] since we have that [itex] F = \psi' (F') [/itex] ... but presumably I did something 'illegal' ... can you explain my error?

    Problem 2 ... How then does it follow that [itex] \phi ( F (d) )) = \phi ( \psi ( F' (d))) [/itex] for any [itex]d \in D[/itex] (given that [itex] F = \psi' (F') [/itex]? Can you help?

    Peter
     
  5. May 12, 2014 #4

    micromass

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    If you simply substituted, then you would have gotten

    [tex]F(d) = \psi^\prime(F^\prime)(d)[/tex]

    Thus ##\psi^\prime(F^\prime)## acts on ##d##. You would not get [tex]F(d) = \psi^\prime(F^\prime(d))[/tex]

    Now, by definition, we have ##\psi^\prime(F^\prime) = \psi\circ F^\prime##. Thus

    [tex]F(d) = \psi^\prime(F^\prime)(d) = (\psi\circ F^\prime)(d) = \psi(F^\prime(d))[/tex]
     
  6. May 12, 2014 #5
    Thanks ... appreciate your help ... given me the confidence to go forward from there in my attempt to understand projective modules

    Thanks again,

    Peter
     
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