Let g: $\mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{Z} \times \mathbb{Z}$ be given by: $g(x,y) = (x+y, x-y)$

Solution 1:

To show $g$ is not a bijection: as you intuited, we can show $g$ is not surjective. Suppose $g(x,y)=(1,0)$. Then for some $x,y\in\Bbb Z$, $x+y=1$ and $x-y=0$. The latter implies $x=y$, so $x+y=2x=1$, which is impossible since $x\in\Bbb Z$.

Finding the image of $g$: first you have to translate the visual pattern you found into set notation. It looks like the image consists of all $(x,y)$ such that, when $y$ is even, $x$ is even, and, when $y$ is odd, $x$ is odd. In other words, $$\operatorname{im}(g)=\{(2x,2y):x,y\in\Bbb Z\}\cup\{(2x+1,2y+1):x,y\in\Bbb Z\}.$$ Once you've written this, you can directly prove it. To prove $A=B$, you must show that $p\in A\implies p\in B$ and $p\in B\implies p\in A$. In this case, $A=\operatorname{im}(g)$ and $B=\{(2x,2y),(2x+1,2y+1):x,y\in\Bbb Z\}$.

• Suppose $p\in A$. Then we can write $p=g(x,y)$ for some $x,y\in\Bbb Z$. That means $p=(x+y,x-y)$. Consider the two possible cases here: either $x$ and $y$ have the same parity, or they have different parities. Show that, in either case, $(x+y,x-y)$ is in $B$.
• Suppose $p\in B$. There are two cases. I will consider the first, and you can fill in the rest. If $p=(2x,2y)$ for some $x,y\in\Bbb Z$, then $p=g(x,x)+g(y,-y)=g(x+y,x-y)\in\operatorname{im}(g)$.