Am I determining the Image correctly? basically we have
$f: \mathbb{Z} \rightarrow\mathbb{N_0} $
with
$f (n) = \left\{
\begin{array}{ll}
-2n & n \leq 0 \\
2n-1 & \, \textrm{n > 0} \\
\end{array}
\right. $
And I should determine $Im(2\mathbb{Z})$ the image of f under $2\mathbb{Z}$ with $2\mathbb{Z} := \{2x | x\in \mathbb{Z}\}$.
(case 1) $n \leq 0$
We put in numbers like $\{...,-8,-6,-4,-2,0 \}$ So $f(-8) = 16, f(-6) = 12, f(-4) = 8, f(-2) = 4, f(0) = 0$
So we have $Im(2\mathbb{Z}) = \{...,16,12,8,4,0\}$ in the case $n \leq 0$
You can see that this set is basically multiplies of $4$ so it can be written as $4n$.
(case 2) $n > 0$
We put in numbers like $\{2,4,6,8,... \}$ So $f(2) = 3, f(4) = 7, f(6) = 11, f(8) = 15$
So we have $Im(2\mathbb{Z}) = \{3,7,11,15,...\}$ in the case $n > 0$
You can see that this set is basically multiplies of $4$ subtracted by 1 so it can be written as $4n-1$.
Can we conclude that $Im(2\mathbb{Z}) = \{  4n \lor 4n-1 \forall n \in \mathbb{N_0}\}$ ?
 A: We want to proof equality of sets. You can seperate cases. Then for $n>0$ you have $2\mathbb{Z}=2\mathbb{N}$, and for $n\leq 0$ we have $-2\mathbb{Z}=\mathbb{N}_0$. So in the first case, try to proof $f(2\mathbb{N})=\{4n-1;n\geq 1\}$. For equality of sets you have to proof "$\subseteq$" and "$\supseteq$".
So lets proof $f(2\mathbb{N})\subseteq \{4n-1;n\geq 1\}$.
For that we have to proof that every $m\in f(2\mathbb{N})$ has the form $4n-1$ where $n$ is a natural number strict greater $0$.
We have $m\in f(2\mathbb{N})$. So there is an $x\in2\mathbb{N}$ with $f(x)=m$. So $x$ is even, and we can write $x=2i$. Furthermore we know that $2i>0$. So $f(2i)=2(2i)-1=4i-1$.
Which is exactly what we had to show. So for $i=n$, we have proved that $m=4n-1$.
The other set equality can then be proved similar.
Also there should be a more direct proof, but splitting it up like this might be helpful at the beginning.
This might look intimidating, as this is a little bit "back and forth", but it is kinda clear what we have to do in each step.
Now try and proof $\supseteq$ similar to the given proof.
Start with: Let $4n-1\in\{4n-1; n\geq 1\}$. Then we have to find a $m\in2\mathbb{N}$ with $f(m)=4n-1$.
