# NeedEditingSpace

## Content

Proof that [ilmath]T[/ilmath] 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 [ilmath]t\in T[/ilmath] be given - Caution:The result requires that [ilmath]T[/ilmath] is empty, but we can still do this[Note 1]
• 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 2]
• 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][Note 3]
• 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]

## Notes

1. TODO: Investigate
2. This is what evenly covered means, each sheet of the covering is homeomorphic to [ilmath]U[/ilmath] via the restriction onto its image of the covering map
3. As the injection page describes, this is equivalent to and is sometimes given as the following form:
• [ilmath]\forall a,b\in V_\beta[q(a)\eq q(b)\implies a\eq b][/ilmath]
This is just the contrapositive of what we have