A bijection is a function that is both one-to-one and onto.
I need to find such a function that maps $\{ -2, -1 \} \cup [0, 1] \rightarrow (0, 1)$ or equivalently, $ (0, 1) \rightarrow \{-2, -1 \} \cup [0, 1]$.
This problem is very confusing because if I exclude any function output for $x \in [0, 1]$ (output occupied by $\{ -1, -2 \}$) I have to find another function output in $(0, 1)$ that associates with $x$, even though all numbers in $(0, 1)$ have already been covered.
Help needed.
We can compose bijections. First let's get a bijection from $[0,1]\to(0,1)$.
To do it, we can consider the following function (see this answer): $f(x) = \left\{ \begin{array}{1 1} \frac{1}{2} & \mbox{if } x = 0\\ \frac{1}{2^{n+2}} & \mbox{if } x = \frac{1}{2^n}\\ x & \mbox{otherwise} \end{array} \right.$
So we have the bijection $f:[0,1]\to(0,1)$. Now consider $g:[0,1]\cup\{-1,-2\}\to[0,1]$ defined by $g(x) = \left\{ \begin{array}{1 1} f(x) & \mbox{if } x \in[0,1]\\ 0 & \mbox{if } x = -1\\ 1 & \mbox{if } x=-2 \end{array} \right.$
Then $h=f\circ g:[0,1]\cup\{-1,-2\}\to(0,1)$ is a bijection.