1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted