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 $\mathbb{R}\to\mathbb{Z}$.
 A: 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.
A: $f^{-1}(T)$ 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.
A: 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 $f^{-1}(T)=\{x\in \mathbb R:f(x)\in T\}=\{x\in \mathbb R:f(x)=2\}=\{x\in \mathbb R:\lfloor\frac{x-2}{2}\rfloor=2\}=\\=\{x\in \mathbb R:2\leq\frac{x-2}{2}<3\}=\{x\in \mathbb R:4\leq x-2<6\}=\\=\{x\in \mathbb R:6\leq x<8\}.$
