A topological property of shapes like $\bot$ in $\Bbb{R}^2$ Let $X$ be a shape like $\bot$ as a subspace of $\Bbb{R}^2$. Is it possible to construct a continuous function $f : X\times X \longrightarrow X$ such that for any $v,w \in X$, $f(v,w) = f(w,v) \in \{v,w\}$ ?
 A: It seems the following. 
For simplicity we shall use the following description. Consider three segments $[0,1]\times\{1\}$,
$[0,1]\times\{2\}$, and  $[0,1]\times\{3\}$. We identify points $(0,1)$, $(0,2)$ and $(0,3)$ to a point $0^*$ and denote the obtained space as $X$. Put $X_i=(0,1]\times\{i\}$ for every $i$. 
We claim that there is no continuous function $f$ satisfying conditions of the question. Indeed, assume the converse. Let $i,j$ be arbitrary non-equal integers from $1$ to $3$. Since the space $X_i\times X_j$ is connected,  $f(X_i\times X_j)\subset X_i\cup X_j$, and the space $X_i\cup X_j$ is a disjoint union of its clopen (that is closed and open) subspaces $X_i$ and $X_j$, we see that there exists a number $k\in\{i,j\}$ such that $f(X_i\times X_j)\subset X_k$. Let $\{i,j\}\setminus \{k\}=\{l\}$. Therefore for all $0<x,y<1$ we have $f((x,k),(y,l))=(x,k)$. The continuity of the function $f$ implies that $f((x,k),0^*)=(x,k)$ and $f(0^*,(y,l))=0^*$. Consider now a function $g:X\to X$ defined as $g(x)=f(x,0^*)$. Then $g|X_k=\operatorname{id}$ and $g|X_l\equiv 0^*$. Let $\{1,2,3\}\setminus \{k\}=\{i’,j’\}$. Similarly to the above we can show that there exists a number $k’\in\{i',j'\}$ such that $g|X_{k’}=\operatorname{id}$. Again similarly to the above we can show that there exists a number $k’’\in\{k,k’\}$ such that $g|X_{k''}\equiv 0^*$, a contradiction, because $g|X_k=\operatorname{id}$ and $g|X_{k'}=\operatorname{id}$.
