Mapping

Credit to @Guillermo Templado
Answer: both.
Proof:

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Proof of injectivity:
$F$ can be written as $F(x)=(\lfloor x \rfloor, x-\lfloor x \rfloor)$
$\;\;\;\;F(x)=F(y)$
$\Rightarrow (\lfloor x\rfloor, x-\lfloor x\rfloor)=(\lfloor y\rfloor, y-\lfloor y\rfloor)$
$\Rightarrow \huge \{ \normalsize \begin{matrix}\lfloor x\rfloor=\lfloor y\rfloor \\ x-\lfloor x\rfloor = y-\lfloor y\rfloor\end{matrix}$
$\Rightarrow x=y$ (by summing the two equations)

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Proof of surjectivity:
$\forall(a,b)\in\mathbb Z\times [0,1),\exists x=a+b\in\mathbb R\;s.t.\;F(x)=(a,b)$

Beautiful problem +1 and solution too. :-) without proof writing, just with the help of definitions of floor function and fractional part of the function, one can easily conclude the given mapping is bijective. (One can see $1$ and higher values are, not included in $[0,1)$ and that's what we want! For surjectivity of fractional part of function)