NeedEditingSpace

Purpose

• For use with the Unique lifting property page

Content

Proof that [ilmath]T[/ilmath] is an open set:[math]\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}} }[/math][math]\newcommand{\udot}{\cup\mkern-12.5mu\cdot\mkern6.25mu\!}[/math][math]\require{AMScd}\newcommand{\d}[1][]{\mathrm{d}^{#1} }[/math]

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