Exercises:Mond - Topology - 2/Section B/Question 5

Section B

Question 5

Is the quotient map from [ilmath][0,1]\times[0,1]\subset\mathbb{R}^2[/ilmath] to the real projective plane, [ilmath]\mathbb{RP}^2[/ilmath], an open map?

Solution

We will consider the square as [ilmath]I^2\subset\mathbb{R}^2[/ilmath] where [ilmath]I:=[0,1]:=\{x\in\mathbb{R}\ \vert\ 0\le x\le 1\}\subset\mathbb{R}[/ilmath] with [ilmath]v_0=(0,0)[/ilmath], [ilmath]v_1=(1,0)[/ilmath], [ilmath]v_2=(0,1)[/ilmath] and [ilmath]v_3=(1,1)[/ilmath].

Definitions

• Let [ilmath]\pi:I^2\rightarrow\mathbb{RP}^2[/ilmath] be the quotient map the question talks about.
• Let [ilmath]\epsilon\in (0,\frac{\sqrt{2} }{2})\subset\mathbb{R} [/ilmath] be given, the upper bound is chosen so the open ball considered at a vertex does not cross any diagonals of the square.
• Let [ilmath]X:=B_\epsilon(v_3)\cap I^2[/ilmath], where the open ball [ilmath]B_\epsilon(v_3)[/ilmath] of radius [ilmath]\epsilon[/ilmath] centred at [ilmath]v_3[/ilmath] is considered in [ilmath]\mathbb{R}^2[/ilmath], thus [ilmath]B_\epsilon(v_3)\cap I_2[/ilmath] is open in the subspace topology [ilmath]I^2[/ilmath] inherits from [ilmath]\mathbb{R}^2[/ilmath]

Outline of solution

We will show that [ilmath]\pi[/ilmath] is not an open map, by showing that [ilmath]\pi(X)[/ilmath] has a boundary point, namely [ilmath]\pi(v_3)[/ilmath] itself and combine this with:

• A set is an open set if and only if it contains non of its boundary points

As [ilmath]\pi(X)[/ilmath] has a boundary point contained in [ilmath]\pi(X)[/ilmath] it cannot be open! We have exhibited an open set of [ilmath]I^2[/ilmath] (namely [ilmath]X[/ilmath]) which is mapped to a non-open set (namely [ilmath]\pi(X)[/ilmath]), thus [ilmath]\pi[/ilmath] cannot be an open map

Solution body

• Let [ilmath]\epsilon>0[/ilmath] be given such that [ilmath]\epsilon\in\mathbb{R} [/ilmath] and [ilmath]\epsilon<\frac{\sqrt{2} }{2} [/ilmath]
  • Define [ilmath]X:=B_\epsilon(v_3;\mathbb{R}^2)\cap I^2[/ilmath] - the intersection of the open ball of radius [ilmath]\epsilon[/ilmath], in [ilmath]\mathbb{R}^2[/ilmath] and [ilmath]I^2[/ilmath], by definition of the subspace topology [ilmath]X[/ilmath] is open in [ilmath]I^2[/ilmath]
    • We claim that [ilmath]\pi(v_3)\in\partial\pi(X)[/ilmath] - that is that [ilmath]\pi(v_3)[/ilmath] is a boundary point of [ilmath]\pi(X)[/ilmath].
    • To show this we will use a point is in the boundary of a set if and only if every open neighbourhood to that point contains both a point in the set and a point not in the set
      • Let [ilmath]U[/ilmath] be an arbitrary open neighbourhood in [ilmath]\mathbb{RP}^2[/ilmath] to [ilmath]\pi(v_3)[/ilmath] be given.
        • As [ilmath]\pi[/ilmath] is continuous we have [ilmath]\pi^{-1}(U)[/ilmath] is open in [ilmath]I^2[/ilmath]
        • As [ilmath]\pi(v_3)\in U[/ilmath] (by definition) and [ilmath]\pi(v_3)=\pi(v_0)[/ilmath] we see from [ilmath]\pi(v_0)\in U[/ilmath] that [ilmath]v_0\in\pi^{-1}(U)[/ilmath]
          • As [ilmath]\pi^{-1}(U)[/ilmath] is an open set and [ilmath]v_0,v_3\in\pi^{-1}(U)[/ilmath] we see there exist [ilmath]\delta_1',\delta_2'>0[/ilmath] such that [ilmath]B_{\delta_1'}(v_0)\subset\pi^{-1}(U)[/ilmath] and [ilmath]B_{\delta_2'}(v_3)\subset\pi^{-1}(U)[/ilmath]
          • Define [ilmath]\delta_1:=\text{Max}(\{\delta_1',\frac{\sqrt{2} }{2}\})[/ilmath] and [ilmath]\delta_1:=\text{Max}(\{\delta_1',\frac{\sqrt{2} }{2}\})[/ilmath] - this will prevent the following balls from crossing the diagonal of the square and keeps them disjoint.
          • Notice the following: [ilmath]B_{\delta_1}(v_0)\subseteq B_{\delta_1'}(v_0)\subset \pi^{-1}(U)[/ilmath] and [ilmath]B_{\delta_2}(v_3)\subseteq B_{\delta_2'}(v_3)\subset \pi^{-1}(U)[/ilmath]
            • Define [ilmath]p_1\in B_{\delta_1}(v_0)[/ilmath] to be any point in [ilmath]B_{\delta_1}(v_0)[/ilmath] such that [ilmath]p_1\ne v_0[/ilmath] and both the [ilmath]x[/ilmath] and [ilmath]y[/ilmath] coordinates are not [ilmath]0[/ilmath] (for example: [ilmath]p_1:=(\frac{\delta_1}{2},\frac{\delta_1}{2})[/ilmath] would do nicely)
            • Define [ilmath]p_2\in B_{\delta_2}(v_3)[/ilmath] to be any point in [ilmath]B_{\delta_2}(v_3)[/ilmath] such that [ilmath]p_2\ne v_3[/ilmath] and both the [ilmath]x[/ilmath] and [ilmath]y[/ilmath] coordinates are not [ilmath]1[/ilmath] (for example: [ilmath]p_2:=(1-\frac{\delta_2}{2},1-\frac{\delta_2}{2})[/ilmath] would do nicely)
              • Notice:
                1. [ilmath]\pi(p_1)\in U[/ilmath] and [ilmath]\pi(p_2)\in U[/ilmath]
                2. [ilmath]\pi(p_2)\in\pi(X)[/ilmath]
                3. [ilmath]\pi(p_1)\notin\pi(X)[/ilmath]
                   • PROVE THIS. It follows from [ilmath]\pi[/ilmath] being injective on [ilmath](0,1)\times (0,1)\subset [0,1]\times[0,1][/ilmath]