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

Section B

Question 5

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

Solution

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

Definitions

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

Outline of solution

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

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

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

Solution body

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