Find homomorphism where $\ker(\phi) \cong \mathbb{Z}^3$

Question

I need to find a homomorphism where $\ker(\phi) \cong \mathbb{Z}^3$ with $$\phi:\mathbb{Z}^4\rightarrow\mathbb{Z}^2$$

I have set $\phi(x,y,u,v)=(x,x)$

Is this a correct answer?

Answer 1

Yes, your answer is correct, since $${\rm ker}(\phi) = \{(0,y,u,v);u,v,y \in \mathbb{Z}\}$$

Simply put $\psi(0,y,u,v) = (y,u,v)$, then $\psi$ is isomorphism between ker$(\phi)$ and $\mathbb{Z}^3$