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Homework Help: Short Exact Sequences

  1. Sep 15, 2010 #1
    Hello! I have just another problem I can't figure out how to solve:

    1. The problem statement, all variables and given/known data
    Consider a homomorphism of short exact sequences (it's all vector spaces):
    [PLAIN]http://img814.imageshack.us/img814/9568/seq.png [Broken]

    Prove that:
    (1) [tex]\sigma[/tex] is surjective iff [tex]\rho[/tex] is injective.
    (2) [tex]\sigma[/tex] is injective iff [tex]\rho[/tex] is surjective.

    2. Relevant equations
    Earlier parts of the exercise:
    (1) [tex]\psi_2 (\hbox{Im } \sigma)=\hbox{Im } \tau[/tex]
    There was also another,
    (2) [tex]\phi_1(\ker \rho)=\ker \sigma[/tex],
    but this is wrong, I'm afraid.

    3. The attempt at a solution
    I'm deeply sorry, but I have no idea where to start.

    Thanks in advance for any hints!
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 15, 2010 #2
    I could be wrong since my algebra was a long time ago, but it seems to me that both of these statements are false.

    Let [tex]k[/tex] be the field over which the vector spaces are taken. To see that (1) is false, let [tex]F_1 = k, E_1 = k, G_1 = 0, F_2 = k, E_2 = k^2, G_2 = k; \phi_1 = \mathrm{id}, \psi_1 = 0, \phi_2 = \iota_1[/tex] is the inclusion along the first axis, [tex]\psi_2 = \pi_2[/tex] is the projection along the second axis; [tex]\rho = \mathrm{id}, \sigma = \iota_1, \tau = 0[/tex]. This diagram commutes since [tex]\phi_2 \rho = \sigma \phi_1 = \iota_1, \psi_2 \sigma = \tau \psi_1 = 0[/tex], and the sequences are exact since [tex]\ker\psi_1 = \mathop{\mathrm{im}}\phi_1 = k[/tex], [tex]\ker\psi_2 = \mathop{\mathrm{im}}\phi_2 = k \times 0[/tex]. Here [tex]\rho[/tex] is bijective, but [tex]\sigma[/tex] is not surjective.

    To see that (2) is also false, exchange the roles of the two rows, and let [tex]\rho = \mathrm{id}, \sigma = \pi_1, \tau = 0[/tex]. Then [tex]\rho[/tex] is bijective, but [tex]\sigma[/tex] is not injective.
  4. Sep 16, 2010 #3
    Thanks, that's what I was afraid of. Too bad I quickly gave up my attempts to construct counterexample. Thank you for you time, it's perfectly clear now.
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