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

For any function $f:X\longrightarrow Y$ and any $K\subseteq Y, \ \ f^{-1}(K)$ is by definition the set $\{x\in X:f(x)\in K\}\subseteq X$. In case $f$ is one to one then for $|K|=1$ (one element set) $\Rightarrow|f^{-1}(K)|\leq1.$
In your case $T=\{2\}$ so $f^{-1}(T)=\{x\in \mathbb R:f(x)\in T\}=\{x\in \mathbb R:f(x)=2\}=\{x\in \mathbb R:\lfloor\frac{x-2}{2}\rfloor=2\}=\\=\{x\in \mathbb R:2\leq\frac{x-2}{2}<3\}=\{x\in \mathbb R:4\leq x-2<6\}=\\=\{x\in \mathbb R:6\leq x<8\}.$
