(May 2013) Define an equivalence relation \(\sim\) on \(\mathbb R^2\) by \((x_1,y_1)\sim(x_2,y_2)\) if and only if \(x_1^2+y_1^2=x_2^2+y_2^2\).
Identify the quotient space \(X=\mathbb R^2/{\sim}\) as a familiar space and prove that it is homeomorphic to this familiar space.
Determine whether the natural map \(p\colon\mathbb R^2\to X\) is a covering map. Justify your answer.
Solution
Define \[ r:\mathbb R^2\to[0,\infty), \qquad r(x,y)=\sqrt{x^2+y^2}. \] The fibers of \(r\) are exactly the equivalence classes. Therefore \(r\) factors uniquely through a bijection \[ \bar r:X=\mathbb R^2/{\sim}\ \longrightarrow[0,\infty) \] with \(r=\bar r\circ p\). Since \(p\) is a quotient map and \(r\) is continuous, \(\bar r\) is continuous. Its inverse is \[ [0,\infty)\to X, \qquad t\mapsto p(t,0), \] which is continuous as the composite of the continuous inclusion \(t\mapsto(t,0)\) with \(p\). Thus \[ X\cong[0,\infty). \]
The quotient map \(p\) is not a covering map. For every \(r>0\), the fiber over the class of radius \(r\) is the circle \[ p^{-1}([r])=\{(x,y):x^2+y^2=r^2\}. \] Fibers of a covering map are discrete subspaces of the total space, but this circle is not discrete. Hence \(p\) cannot be a covering map.