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Surjectivity of induced map via hom functor implies injectivity

  1. Apr 11, 2013 #1
    1. The problem statement, all variables and given/known data

    Let R be an arbitrary ring, B and B' be left R-modules, and [itex] i: B' \to B [/itex] be an R-module morphism. Show that if the induced map [itex] i^*: \operatorname{Hom}_R(B,M) \to \operatorname{Hom}(B',M) [/itex] is surjective for every R-module M, then [itex] i: B' \to B [/itex] is injective.

    3. The attempt at a solution

    The maps all seem to go the wrong way to use the categorical definition of kernels, so I fear that I must be much trickier about the application and exploit the module structure quite specifically. This would suggest an intelligent choice of M and a morphism [itex] B' \to M [/itex] in order to apply the hypothesis.

    In my mind, the only obvious candidate is the projection map [itex] \pi: B' \to B'/\ker i [/itex]. The hypothesis would then suggest that there exists [itex] \hat \pi: B \to B'/\ker i [/itex] such that [itex] \hat \pi \circ i = \pi [/itex]. However, nothing useful seems to come of this. Any ideas?
     
  2. jcsd
  3. Apr 11, 2013 #2

    micromass

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    What if you take ##M=B^\prime## and take the identity morphism in ##Hom(B^\prime,B^\prime)##?
     
  4. Apr 12, 2013 #3
    Ah yes excellent. I figured that out today, and could have saved myself a lot of time if I had just looked here first.
     
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