Composition of functions is constant in $\mathbb{R^2}$. Let $\hspace{0.05cm}f:\mathbb{R^2}\to\mathbb{R^2}$ $\hspace{0.05cm}$ and $\hspace{0.05cm}$$g:\mathbb{R^2}\to\mathbb{R^2}$ $\hspace{0.05cm}$  be such that$\hspace{0.05cm}$ $g\circ f$$\hspace{0.05cm}$ is constant,then which of the following must be constant?
$1.f$ $\hspace{0.8cm}$ $2.g$ $\hspace{0.8cm}$ $3.f\circ g$ $\hspace{0.8cm}$ $4$.None 
What i did:For $\hspace{0.05cm}f:\mathbb{R}\to\mathbb{R}$ $\hspace{0.05cm}$ and $\hspace{0.05cm}$$g:\mathbb{R}\to\mathbb{R}$ $\hspace{0.05cm}$
I find example $$f(x) = \begin{cases} 1 & x\in\mathbb{Q} \\ -1 &x\notin\mathbb{Q}  \end{cases}$$
$$g(x) = x^2, \,\,\, \forall x \in \mathbb{R}$$
According to this example my answer for $\mathbb{R}$ is option D.
But for $\mathbb{R^2}$ i try to find such an example but i could not find such an example.Please someone help me to solve this.
 A: Let $f(0,0)=(-1,0)$, and $f(x,y)=(1,0)$ if $(x,y)\neq (0,0)$.  Let $g(x,y)=(x^2,0)$.  
A: This answer emphasizes that only the cardinality of $X$ matters.
More generally let $f,g:X\rightarrow X$ be functions. Let's start with the following observation:

If $c:X\rightarrow X$ is a constant function then functions
  like $c\circ f$ and $g\circ c$ are constant for any $f,g$.

A) $X$ is empty or is a singleton.
Then any function $h:X\rightarrow X$ is a constant function. So a 'yes' on questions 1), 2) and 3). The answer on 4) can easily be deduced from the answers on 1), 2) and 3).
B) $X=\{a,b\}$ where $a\neq b$. 
The mentioned observation tells us that the condition that $g\circ f$ is constant does not induce any demands on $f$ or $g$ separately. Secondly non-constant functions $X\rightarrow X$ exist here. So the answer on questions 1) and 2) is 'no'. But in this situation we do have the implications: $$g\circ f\text{ constant}\Rightarrow f\text{ constant}\vee g\text{ constant}\Rightarrow f\circ g\text{ constant}$$
So here a 'yes' on 3).
C) Set $\{a,b\}$ where $a\neq b$ is a proper subset of $X$. 
Also here off course 'no' on questions 1) and 2). Also now a 'no' on 3). Let $f$ satisfy $f\left(x\right)=b$ if $x\neq a$ and $f\left(a\right)=a$.
Let $g$ satisfy $g\left(x\right)=x$ if $x\notin\left\{ a,b\right\} $
and $g\left(a\right)=g\left(b\right)=a$. Then $g\circ f$ is constant
but $f\circ g$ is not.

It is clear that we are in case C) if  $X=\mathbb{R}^{2}$. 
