Bijection from $\Bbb N\to \Bbb Z \times \{1, 2, 3, 4\}$. Find a bijection from $\Bbb N\to \Bbb Z \times \{1, 2, 3, 4\}$.
Ok, so I know some element $x$, in $\Bbb N$ maps to an element $(y,z)$ in $\Bbb Z.$
I know to to get from $x$ to $y.$
But since $z$ can only be $1,2, 3,$ or $4$, I'm lost.
 A: Assuming $\mathbb{N} = \{0,1,2,3,...\}$, I will write two bijections.


*

*Let $f: \mathbb{N} \rightarrow \mathbb{Z}$ be this function.  $f(0) = 0$.  $f(1) = 1$,  $f(2) = -1$.  $f(3) = 2$.  $f(4) = -2$.  Etc.  This function is surjective because if I am given $z \in \mathbb{Z}$ is positive, we had $f(2z - 1) = z$.  If I am given $z \in \mathbb{Z}$ is negative, we had $f(-2z) = z$.  Also $f(0) = 0$.  This function is injective because if $z_{1} = z_{2} \in \mathbb{Z}$, then under any of the three cases we are in ($2z-1$, $-2z$, or $0)$, these expressions coincide when we plug in $z_{1} = z_2$.  Since the function is both surjective and injective, the function is bijective.

*Let $g: \mathbb{Z} \rightarrow \mathbb{Z} \times \{1,2,3,4\}$ be this function.  The Euclidean algorithm says $z = 4q + m$ where $q \in \mathbb{Z}$ and $m \in \{1,2,3,4\}$ and this choice of $(q,m)$ is unique once we have decided we take $m \in \{1,2,3,4\}$. 
The bijection is $g(z) = (q,m)$.  This function is surjective because if I am given $(q,m)$, we had $g(4q+m) = (q,m)$.  This function is injective because if I am given $q_1 = q_2 \in \mathbb{Z}$ and $m_{1} = m_{2} \in \{1,2,3,4\}$, then $4q_{1} + m_{1} = 4q_{2} + m_{2}$.
Both maps are bijective, so their composition is bijective.
A: The idea is to take a bijection from $\Bbb N$ to $\Bbb Z$ then "spread it evenly amongst the four equivalence classes modulo four". This can be done since the cardinality of $\Bbb N$ and $\Bbb Z^4$ is the same.

Define $\xi:\Bbb N\to \Bbb Z\times \{1,2,3,4\}$ by some ordinary map $\mu$ from $\Bbb N$ to $\Bbb Z$, like one from here and considering the input modulo four: $$\xi: n=4k+m\mapsto (\mu(k), [m]_4),$$ where $$[m]_4:=\{\ell\mid \ell=4h+m\text{ for some } h\in \Bbb N\}$$ and be sure to identify such an equivalence class with its representative in $\{1,2,3,4\}$, so $[0]_4$ is identified with $4$.

One could also argue for the existence of such a bijection alone by writing $$\Bbb Z\times \{1,2,3,4\}\cong \bigcup_{i=1}^4 \Bbb Z\times\{i\}$$ and saying that a countable union of countable sets is countable, hence a bijection with $\Bbb N$ exists.
A: Define $\sigma: \Bbb Z \times \{1, 2, 3, 4\} \to  \Bbb Z \times \{1, 2, 3, 4\}$ by
$$   
    \sigma(z,k) = \left\{\begin{array}{lr}
        (z, k+1), & \text{for } z = 0 \, \land \, k \lt 4\\
        (-1, 1), & \text{for } z = 0 \, \land \, k = 4\\
        (z, k+1), & \text{for } |z| \gt 0 \, \land \, k \lt 4\\
        (-z, 1), & \text{for } z \lt 0  \, \land \, k = 4\\
        (-z-1, 1), & \text{for } z \gt 0  \, \land \, k = 4
        \end{array}\right\} 
$$
Exercise: Show that $n \mapsto \sigma^n(0,1)$ is a bijective mapping between $\{0,1,2,3,...\}$ and $\Bbb Z \times \{1, 2, 3, 4\}$.
