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.