Consider the function $f\colon\mathbb{Z}\times\{0,1\}\rightarrow \mathbb{Z}$ defined by: $$ f(n,m)=n+2^m. $$
For injective: Let $f(a,b)=f(c,d)$. Therefore, $a+2^b=c+2^d$.
I am unsure how to prove surjectivity and to how to prove the rest of my injective proof.
On
To prove injectivity, we need to show that $f(a,b) = f(c,d)$ only if a=c and b = d. However, it is not the case as questionned in the comment. Since we can have $f(0,1)$ = $f(1,0)$, where a $\neq$ c and b $\neq$ d!!! So, $f$ cannot be injective by definition.
Regarding surjectivity, we need to ensure that for every z $\in$ $\mathbb{Z}$, we can find a couple
$(a,b)$ $\in$ $\mathbb{Z}$ $\times$ $\{$$0,1$$\}$.
So, we need to find a map $\phi$ : $\mathbb{Z}$ $\rightarrow$ $\mathbb{Z}$ $\times$ $\{$$0,1$$\}$.
Hopefully, such map does exist! We need to create a map like this one below.
$$\phi(z)=\begin{cases} (\frac{z}{2},0)&\text{if}\, z\text{ is even } \\ (\frac{(z-1)}{2},1)&\text{if}\, z\text{ is odd}\\ \end{cases}$$
The existence of such map ensure that we can map every element of the domain into a desired couple of the image of the function $f$. Therefore, this shows that the function $f$ is surjective by definition.
Injectivity:
Suppose $a+2^b=c+2^d$ and $b \ne d$. We can assume (symmetry) that $b = 0$ and $d = 1$. Then
$\quad a+1=c+2$
So we have, for example, $f(1,0) = f(0,1) = 2$ and $f$ is not injective.
Surjectivity:
Solve
$\quad a+2^b = n$.
OK,
$\quad f(n-1, 0) = n$
and so $f$ is surjective.