I found the piece-wise bijection:
$f(x)=\begin{cases} 0 & x=1 \\ e^{x-1} & x \in \{2,3,4,...\} \\ e^{x} & x \in \mathbb{R}- \mathbb{N} \end{cases}$
But can we find a non piece-wise bijection?
I tried the composition idea:
$f_{1}:(0,1] \rightarrow [0,\infty)$ and $f_{2}:\mathbb{R} \rightarrow (0,1]$
I know $f_{1}=\text{sech}^{-1}(x)$ but I couldn't find a non piece-wise bijection $f_{2}$.
Is there a non piece-wise bijection between $\mathbb{R}$ to $[0,\infty)$?
As @PaulSinclair commented, I suppose you are looking for continuous bijection.
Assume $f:\mathbb{R}\to[0,\infty)$ is continuous surjection.
Thus, $\exists x \in \mathbb{R}$ such that $f(x) = 0$. Choose $\epsilon > 0$.
Now,
$f$ is continuous $\implies f([x-\epsilon ,x])= [0,a_1]$ and $f([x, x+ \epsilon ])= [0,a_2]$ for some $a_1,a_2 \in [0, \infty)$.
Let $a = \min \{a_1,a_2\}$
Thus, as $f$ is continuous, $\forall y \in (0,a],\ y$ has two preimages $x_1, x_2$;
$x_1 \in [x-\epsilon, x)\ \text{and } x_2 \in (x, x +\epsilon]$
Thus $f$ can't be one one, and subsequently not a bijection