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 $\mathbb{R}\to\mathbb{Z}$.

$f^{-1}(T)$ could mean either the preimage of $T$ under $f$, or the image of $T$ under $f^{-1}$. The preimage always exists, and is defined to be the set of those points that map into $T$, i.e.
$$f^{-1}(T)=\{x\mid f(x)\in T\}.$$
If $f$ has an inverse, there is also the image of $T$ under $f^{-1}$ which is
$$f^{-1}(T)=\{f^{-1}(y)\mid y\in T\}.$$
However, in this case if we denote the preimage by $A$ and the inverse image by $B$ we get
$$x\in A\implies f(x)\in T\implies x=f^{-1}(f(x))\in B$$
and
$$f^{-1}(y)\in B\implies y=f(f^{-1}(y))\in T\implies f^{-1}(y)\in A$$
so the sets are actually the same.
