Proving Surjectivity and Injectivity in Homomorphisms of Short Exact Sequences

[losiu99]

Hello! I have just another problem I can't figure out how to solve:

Homework Statement

Consider a homomorphism of short exact sequences (it's all vector spaces):
[PLAIN]http://img814.imageshack.us/img814/9568/seq.png

Prove that:
(1) $$\sigma$$ is surjective iff $$\rho$$ is injective.
(2) $$\sigma$$ is injective iff $$\rho$$ is surjective.

Homework Equations

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

The Attempt at a Solution

I'm deeply sorry, but I have no idea where to start.

Thanks in advance for any hints!

 

[ystael]

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

To see that (2) is also false, exchange the roles of the two rows, and let $$\rho = \mathrm{id}, \sigma = \pi_1, \tau = 0$$. Then $$\rho$$ is bijective, but $$\sigma$$ is not injective.

 

[losiu99]

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.