$f:\mathbb{N}\to\mathbb{Z}$ which is surjective but not injective I'm looking for a really simple example of $f:\mathbb{N}\to\mathbb{Z}$ which is surjective but not injective. I thought about:
$$
f(x)=\begin{cases}
0 & x=0\\
\frac{x+1}{2} & x\text{ is odd}\\
-\frac{x}{2} & x\text{ is even}\wedge x\neq0
\end{cases}
$$
Is there a better and more simple  example I can use?
 A: You can take, for instance, $f(n)=(-1)^n\left\lfloor\dfrac n2\right\rfloor$ (assuming, as you did, that $0\in\Bbb N$).
A: $0 \notin \mathbb{N}$.
Let $f:\mathbb{N} \to \mathbb{Z}$ be a surjective and injective function. Define:
$$g(n) = \cases {f(1),\quad n =1\\
                 f(1),\quad n =2\\
                 f(n-1),\quad n\geq3}$$
Then $g:\mathbb{N} \to \mathbb{Z}$ is surjective but not injective.
A: $$
f(x)=\begin{cases}
0 & x=0\\
\frac{x+1}{2} & x\text{ is odd}\\
-\frac{x}{2} & x\text{ is even}\wedge x\neq0
\end{cases}
\implies
f(n)=(-1)^{n+1}\bigg\lfloor\frac{n+1}{2} \bigg\rfloor
$$
$$
f(0)=(-1)^1\bigg\lfloor\frac{1}{2}\bigg\rfloor=0\quad 
f(1)=(-1)^2\bigg\lfloor\frac{2}{2}\bigg\rfloor=1\quad 
f(2)=(-1)^3\bigg\lfloor\frac{3}{2}\bigg\rfloor=-1\\
f(3)=(-1)^4\bigg\lfloor\frac{4}{2}\bigg\rfloor=2\quad 
f(4)=(-1)^5\bigg\lfloor\frac{5}{2}\bigg\rfloor=-2\quad 
f(5)=(-1)^6\bigg\lfloor\frac{6}{2}\bigg\rfloor = 3$$
Of course, the following alone also meets the criteria: $$f(x)=(-1)^x$$
as does the floor function alone unless you talk about infinities.
