Find a bijection between $[0,1)\sim(0,1)$

Question

Find a bijection between $[0,1)\sim(0,1)$.

Well I know that $$[0,1)\sim(0,1)=[0,\frac{1}{2})\cup [\frac{1}{2},1)\sim(0,\frac{1}{2}]\cup [\frac{1}{2},1).$$

So I found this function: \begin{align} f(x)&=-x+\frac{1}{2} &\text{ if } x\in [0,\frac{1}{2})\\ f(x)&=x &\text{ if } x\in [\frac{1}{2},1)\\ \end{align} Which is surjective but not injective because $f^{-1}(\frac{1}{2})=0,\frac{1}{2}$. Is this the correct conclution to make about $f(x)$? What is a function that works/is bijective?

This question is different than the other questions on bijections between simple real intervals because I want feedback on this specific function, $f(x)$, and I find mapping from a half closed set to an open set more confusing than mapping an open set onto a closed set.

Answer 1

It is not a bijection because both $0$ and $\frac 12$ are mapped to $\frac 12$. Mapping between open and closed intervals make you take care of both endpoints. When you have a half open interval and an open one, you only need to take care of one endpoint, but doing two is no harder than doing one. The suggested duplicate and others show an explicit bijection. The trick is to find a series of points and push them all down the line to make room for the one you need to get rid of.