Can floor functions have inverses? R to R
$f(x) = \lfloor \frac{x-2}{2} \rfloor $
If $T = \{2\}$, find $f^{-1}(T)$
Is $f^{-1}(T)$ the inverse or the "image", and how do you know that we're talking about the image and not the inverse?
There shouldn't be any inverse since the function is not one-to-one, nor is it onto since it's $\mathbb{R}\to\mathbb{R}$ and not $\mathbb{R}\to\mathbb{Z}$.

Note to this calculating;
$[\frac{x-2}{2}]=2\Longrightarrow 2\leq\frac{x-2}{2}<3\Longrightarrow 4\leq x-2<6\Longrightarrow 6\leq x<8$
so the set $f^{-1}(\{2\})$ is equal to $[6,8)$.
Now;
$\forall y\in\mathbb{Z}\; :\; f^{-1}(\{y\})=\{x\in\mathbb{R}|[\frac{x-2}{2}]=y\}$ ... $\Longrightarrow\{x\in\mathbb{R}|x\in[2y+2,2y+4)\}$
In final;
$\Longrightarrow \forall T\subset\mathbb{R}\; :\; f^{-1}(T)=\cup_{y\in T\cap\mathbb{Z}}[2y+2,2y+4)$
and if $T\cap\mathbb{Z}=\emptyset$ then $f^{-1}(T)=\emptyset$ too.
And about existence of inverse functions. If a function be one-to-one it has left-inverse and if it be onto it has right-inverse. for existence both it should be bijective. But always we can define a function which bring back any point of range to set of elements that their value by f is them. like what we had done above.
