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Linear applications simple doubt

  1. Oct 22, 2013 #1
    Is an injective endomorfism automatically bijective?
     
  2. jcsd
  3. Oct 22, 2013 #2
  4. Oct 23, 2013 #3
    Thanks for confirming that intuition and for the pointer to Fredholm alternative, never heard of it.
     
  5. Oct 24, 2013 #4

    WWGD

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    Doesn't this follow from rank-nullity theorem?
     
  6. Oct 25, 2013 #5

    mathwonk

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    for finitely generated modules over other rings this can fail, e.g. if Z is the integers, the injective endomorphism Z-->Z taking n to 3n is injective but not surjective. Interestingly however, a surjective endomorphism of a finitely generated module is always injective as well.
     
  7. Oct 25, 2013 #6
    Yes. But the Fredholm alternative also hold in some infinite-dimensional cases. There, you can't use rank-nullity.
     
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