Find kernel of $([x+y]_2, 2x-y)$ In my practice problem I have:
$\phi=\mathbb{Z}\times \mathbb{Z} \rightarrow (\mathbb{Z}/2\mathbb{Z})\times \mathbb{Z}$ by homomorphism
$\phi(x,y)=([x+y]_2, 2x-y)$
Find $\ker(\phi)$ and show it is isomorphic to $\mathbb{Z}$
My answer:
$\ker(\phi)=\{ (x,2x) \in \mathbb{Z}^2 | x \in 2\mathbb{Z} \}$ as this ensures $[x+y]=[0]$ and $2x=y$
Let $f=x$ for $x \in 2\mathbb{Z}$. It is injective because $f=(0)$ iff $x=0$
Is my answer correct and how do I show it is surjective? It seems strange that $\mathbb{Z} \cong 2\mathbb{Z}$ because what would the preimage of $13 \in \mathbb{Z}$ be in $2\mathbb{Z}$
 A: As additive groups we indeed have $\Bbb Z\cong 2\Bbb Z$ by $x\mapsto 2x$, so the correspondent element of $13\in\Bbb Z$ is $26\in2\Bbb  Z$.
Your solution for the kernel is fine, we can write it alternatively as
$$\ker\phi=\{(2k,\, 4k):k\in\Bbb Z\}$$
and then define $f:\ker\phi\to\Bbb Z,\ f((2k,4k))=k$.
If you wish, you can express the same by $f((x,2x))=x/2$ and noting $x\in 2\Bbb Z$ which ensures $x/2\in\Bbb Z$.
Its inverse is $g(k)=(2k,\, 4k)$.
A: I agree with you for the kernel.
We have $\text{ker}(\phi)=\{(2u,4u)|u\in \mathbb{Z}\}$
To find an isomorphism, we can for example consider the application::
$\varphi: \mathbb{Z}\to \text{ker}(\phi)$ such that $\varphi(x)=(2x,4x)$
I pass on the check of homomorphism. The injectivity is clear...
Regarding surjectivity, it's pretty obvious, but you have to write it down.
Let $f\in \text{ker}(\phi)$. By definition of $\text{ker}(\phi)$, there exists $u\in\mathbb{Z}$ such that $f=(2u,4u)$ and we remark that $f=\varphi(u)$.
For your last remark, the isomorphism Between $\mathbb{Z}$ and $2\mathbb{Z}$ (quite natural) is the multiplication by 2 and the surjectivity comes from the fact that we find an antecedent of an element of $2\mathbb{Z}$ by dividing by 2.
