1. ## Diagram

Hi every one, I have this diagram with isomorphisms, which I can't get to commutate:

the texts says that the horizontal homomorphisms are the obvious ones (whatever is meant by that) and I am to find 3 nonzero vertical homomorphisms so that the diagram commutates. Also to find the kernel-cokernel sequence.

So if I say that
$\displaystyle f_1:2\mathbb{Z}\to \mathbb{Z}$ by $\displaystyle f_1=1/2 z$
$\displaystyle f_2:\mathbb{Z}\to \mathbb{Z}/2\mathbb{Z}$ by $\displaystyle f_2=[z]_2$
$\displaystyle g_1:4\mathbb{Z}\to \mathbb{Z}$ by $\displaystyle g_1=1/4 z$
$\displaystyle g_2:\mathbb{Z}\to \mathbb{Z}/4\mathbb{Z}$ by $\displaystyle g_2=[z]_4$

and then define
$\displaystyle \phi_1: 2\mathbb{Z}\to 4\mathbb{Z}$ by $\displaystyle \phi_1=2z$
$\displaystyle \phi_2: \mathbb{Z}\to \mathbb{Z}$ by $\displaystyle \phi_2=z$
$\displaystyle \phi_3: \mathbb{Z}/2\mathbb{Z}\to \mathbb{Z}/4\mathbb{Z}$ by $\displaystyle \phi_3=[z]_4$

then I can't get the right square to cummutate, for if I take f.eg. z=3 then $\displaystyle g_2\phi_2(3)=-1\neq 1=\phi_3f_2(3)$

How can i get it right?

the kernel would be (I guess?)
$\displaystyle 0\to Ker\phi_1\to Ker\phi_2\to Ker\phi_3 \to CoKer\phi_1 \to CoKer\phi_2 \to CoKer\phi_3 \to 0$ because the rows are exact.

2. Originally Posted by stephi85
Hi every one, I have this diagram with isomorphisms, which I can't get to commutate:
Try this one, and let me know if it is not working.

$\displaystyle f_1:2\mathbb{Z}\to \mathbb{Z}$ by $\displaystyle f_1(z)=z$
$\displaystyle f_2:\mathbb{Z}\to \mathbb{Z}/2\mathbb{Z}$ by $\displaystyle f_2(z)=[z]_2$
$\displaystyle g_1:4\mathbb{Z}\to \mathbb{Z}$ by $\displaystyle g_1(z)=z$
$\displaystyle g_2:\mathbb{Z}\to \mathbb{Z}/4\mathbb{Z}$ by $\displaystyle g_2(z)=[z]_4$

and then define
$\displaystyle \phi_1: 2\mathbb{Z}\to 4\mathbb{Z}$ by $\displaystyle \phi_1(z)=2z$
$\displaystyle \phi_2: \mathbb{Z}\to \mathbb{Z}$ by $\displaystyle \phi_2(z)=2z$
$\displaystyle \phi_3: \mathbb{Z}/2\mathbb{Z}\to \mathbb{Z}/4\mathbb{Z}$ by $\displaystyle \phi_3(z)=[2z]_4$

Then, $\displaystyle g_2\phi_2(3) = \phi_3f_2(3)= [2]_4$ .

the kernel would be (I guess?)
$\displaystyle 0\to Ker\phi_1\to Ker\phi_2\to Ker\phi_3 \to CoKer\phi_1 \to CoKer\phi_2 \to CoKer\phi_3 \to 0$ because the rows are exact.
This seems like a snake lemma. To apply a snake lemma, the necessary conditions are

f_1 is a monomorphism and g_2 is an epimorphism.

So I think we can apply a snake lemma here.

3. Seems like that one actually did the trick.
Thanks a lot!