NeedEditingSpace

Purpose

• For use with the Unique lifting property page

Content

Proof that $T$ is an open set: $$\newcommand{\bigudot}{ \mathchoice{\mathop{\bigcup\mkern-15mu\cdot\mkern8mu}}{\mathop{\bigcup\mkern-13mu\cdot\mkern5mu}}{\mathop{\bigcup\mkern-13mu\cdot\mkern5mu}}{\mathop{\bigcup\mkern-13mu\cdot\mkern5mu}} }$$ $$\newcommand{\udot}{\cup\mkern-12.5mu\cdot\mkern6.25mu\!}$$ $$\require{AMScd}\newcommand{\d}[1][]{\mathrm{d}^{#1} }$$

• Let $t\in T$ be given - Caution:The result requires that $T$ is empty, but we can still do this
  • Define $r_1:\eq g(t)$, note $r_1\in E$
    • Define $r_2:\eq h(t)$, note $r_2\in E$
      • Define $z:\eq p(r_1):\eq p(g(t))\eq f(t)$ by $g$ being a lifting, note $z\in X$
        • Note that $z:\eq p(g(t))\eq f(t)\eq p(h(t))\eq: p(r_2)$ by $h$ being a lifting, so we have
          • $p(r_1)\eq p(r_2)$
        • As $p$ is a covering map we see that:
          • $\exists U\in\mathcal{J}[z\in U\wedge U\text{ is evenly covered by }p]$
            • By evenly covered: $\exists(V_\alpha)_{\alpha\in I}\subseteq\mathcal{H}\big[\bigudot_{\alpha\in I}V_\alpha\eq p^{-1}(U)\wedge \big(\forall\beta\in I[V_\beta\cong_{p\vert_{V_\beta}^\text{Im} } U]\big)\big]$ (note that $\bigudot$ emphasises the union is of pairwise disjoint sets)
        • Define $(V_\alpha)_{\alpha\in I} \subseteq H$ to be the arbitrary family of sets open in $E$ which are the sheets of the covering of $U$.
          • Then:
            1. $\exists\beta\in I[r_1\in V_\beta]$
            2. $\exists\gamma\in I[r_2\in V_\gamma]$
          • Lemma: $\beta\neq\gamma$
            • Suppose that $\beta\eq\gamma$ (and thus use $V_\beta$ as the notation and discard $\gamma$)
              • Then $r_1\in V_\beta$ and $r_2\in V_\beta$
              • Define $q:\eq p\vert^\text{Im}_{V_\beta}:V_\beta\rightarrow U$, so $V_\beta$$\cong$${}_q U$
                • Note that as it is a homeomorphism it is a bijection and as it is a bijection it is an injection, so:
                  • $\forall a,b\in V_\beta[a\neq b\implies q(a)\neq q(b)]$
                • So $r_1\neq r_2\implies q(r_1)\neq q(r_2)$
                  • But $q(r_1)\eq p(r_1)\eq p(r_2)\eq q(r_2)$ (from above) - so $q(r_1)\eq q(r_2)$
                    • This is a contradiction, so we cannot have $\beta\eq\gamma$
            • Thus $\beta\neq\gamma$
          • By the lemma and the property of the covering sheets (that they're pairwise disjoint) we see:
            • $V_\beta\cap V_\gamma\eq\emptyset$ (they're disjoint)
          • Define $q_\beta:\eq p\vert^\text{Im}_{V_\beta}:V_\beta\rightarrow U$ so that $V_\beta\cong_{q_\beta} U$
            • Define $q_\gamma:\eq p\vert^\text{Im}_{V_\gamma}:V_\gamma\rightarrow U$ so that $V_\gamma\cong_{q_\gamma} U$
              • Now $V_\beta\cong_{q_\beta} U {}_{q_\gamma}\cong V_\gamma$
              • Define $W_\beta:\eq g^{-1}(V_\beta)$ - by continuity of $g$ and the fact that $V_\beta$ is open we see $W_\beta\in\mathcal{K}$, i.e. that $W_\beta$ is open in $Y$
                • Define $W_\gamma:\eq h^{-1}(V_\gamma)$ - by continuity of $h$ and the fact that $V_\gamma$ is open we see $W_\gamma\in\mathcal{K}$, i.e. that $W_\gamma$ is open in $Y$
                  • Note that $t\in g^{-1}(r_1)$ and $r_1\in V_\beta$, so $t\in W_\beta$
                  • Note also that $t\in h^{-1}(r_2)$ and $r_2\in V_\gamma$, so $t\in W_\gamma$
                  • Define $W:\eq W_\beta\cap W_\alpha$, as we're in a topology the intersection of (finitely) many open sets is again open, so we see $W\in\mathcal{K}$ - i.e. $W$ is open in $Y$.
                    • Notice that $t\in W$ as for an intersection we see $t\in W_\beta\cap W_\gamma\iff(t\in W_\beta\wedge t\in W_\gamma)$
                    • --------------------------------------MARKER---------------------------------------
                    • Notice also that $g(W)\subseteq V_\beta$ and $h(W)\subseteq V_\gamma$
                    • As $V_\beta\cap V_\gamma\eq\emptyset$ we see $h(W)\cap g(W)\eq\emptyset$ -
                      TODO: Make a statement page with this, I've used it before!
                      • Because $h(W)\cap g(W)\eq\emptyset$ they cannot agree anywhere, thus $W\subseteq T$
                        • Since $W$ is open and contains $t$ we have found an open neighbourhood contained within $T$ for each point in $T$