Prove that $f$ is NOT surjective

Question

Let $f: Z \times Z \to Z \times Z$ defined like this: $f(x,y) = (x+y, x-y)$

Prove that $f$ is injective, and not surjective.

For injectivity I did that:

Let $(a,b) \in Z\times Z$ and $(c,d) \in Z\times Z$ such that: $f(a,b) = f(c,d)$

and let's show that: $(a,b) = (c,d)$

$$(a+b,a-b) = (c+d,c-d)$$ $$a+b = c+d $$ and $$a-b = c-d$$

and we end up with: $a=c$ and $b=d$, that concludes that: $(a,b) = (c,d)$ which proves that $f$ is injective.

Can someone show me how to prove that $f$ is NOT surjective? I couldn't do it by myself.

Thanks.

Answer 1

It is not surjective. For example $(1,0)$ dont have a source. Assume that $$(x+y,x-y)=(1,0).$$ Then $x=y$ and then $2x=1$ which have no solutions over $\mathbb{Z}$.

By the way over $\mathbb{R}$ it is surjective since the corresponding matrix is invertible.

Answer 2

Let $\phi$ be a mapping $\mathbb{Z}\times\mathbb{Z} \to \{0,1\}$ defined as $$\phi(x,y) = (x + y) \bmod 2.$$

Observe that $\phi \circ f \equiv 0$, because $$\phi\big(f(x,y)\big) = \big((x+x) + (y-y)\big) \bmod 2 = 0.$$ However, there are pairs $(x,y) \in \mathbb{Z}\times\mathbb{Z}$ for which $\phi(x,y) = 1 \neq 0$, for example $(1,0)$. Therefore, $f$ cannot be surjective.

As for injectivity, you could also prove that $f^{-1}(x,y) = \left(\frac{x+y}{2},\frac{x-y}{2}\right)$.

I hope this helps $\ddot\smile$