analysis - Updated: Constructing a bijection between $left(-frac{pi}{2}, frac{pi}{2}right]$ and $mathbb{R}$

I am supposed to construct a bijective function for the interval:
$$\begin{align} I_2=\left(-\frac{\pi}{2} ,\frac{\pi}{2} \right] \longrightarrow \mathbb{R} \tag{Problem} \end{align}$$

I first tried the easier case, i.e.
$$\begin{align}f_1:I_1=\left(-\frac{\pi}{2} ,\frac{\pi}{2} \right) &\longrightarrow \mathbb{R} \\ x& \longmapsto \tan(x) \end{align}$$
which is a bijection. Now I know that the composition of bijective functions is still a bijection. Which means that it should be possible to 'make room' for the missing point $\pi/2$. The following function:
$$\begin{align}\phi : \mathbb{R} &\longrightarrow \mathbb{R \setminus}\lbrace 0 \rbrace \\ x & \longmapsto \begin{cases}x+1 \ \text{if} \ x \in \mathbb{N}_0 \\x \ \text{otherwise} \end{cases}\end{align}$$
would be bijective, such that the composition $\phi \circ f_1: I_1 \rightarrow \mathbb{R}\setminus \lbrace 0 \rbrace$ is bijective, and as desired it now has room for a point that I can map to.

At this point I am not sure if my approach is correct because I can't find a function that would do the trick. Would I need to come up with another composition or is it enough to define a function that maps to the functions introduced above?

Update (in consideration of the answers given)

If I understand things correctly I can define: $$\begin{align}f_2: I_2 &\longrightarrow \mathbb{R} \\ x& \longmapsto \begin{cases}\phi(x) \ \text{for} \ x \in I_2 \\0 \ \text{for} \ x=\frac{\pi}{2} \end{cases} \end{align}$$
Update 2 (Clarification required).

Define a new function to be equal to $\phi f_1$ over $I_1$ and have it map $π/2$ to $0$. As suggested (and upvoted) by @TBrendle.

If I do understand this correctly, then I need to map $x=\frac{\pi}{2}$ to $0$. However in this case it would make no sense to me to include $I_1$ in the domain, because $\pi/2$ is not in the domain, hence I don't see why I should include it in the codomain, however if I define:
$$\begin{align}w: I_2 &\longrightarrow \mathbb{R} \\ (\phi\circ f_1)(x) & \longmapsto \begin{cases} (\phi \circ f_1) \ \text{if} \ x \in I_2 \\ 0 \ \text{if} \ x= \frac{\pi}{2} \end{cases} \end{align}$$
This doesn't even look like a legitimate function to me anymore, since at $x=0$ the function evaluates to both $0$ and $1$.

Answer

Your approach is flawless. What were you worried about? Your function will be defined piecewise.

Added details:

Your new function is
$$\psi : \left(-\frac{\pi}{2} \frac{\pi}{2}\right] \to \mathbb{R}$$

given by

$$\psi(x) = \begin{cases} \phi(\tan x), & \text{for }-\frac{\pi}{2} < x< \frac{\pi}{2} \\ 0, & \text{for } x=\frac{\pi}{2}\\ \end{cases}$$
where $\phi$ is as given in the question. You can verify that the function $\psi$ has the specified domain and range and is a bijection.