### add version of alternative proof

parent c6c4ed4e
 %! TEX program = lualatex \input{.maindir/tex/header/preamble-section} % inputs the preamble only if necessary \docStart \providecommand*{\ver}{#1 [#2]} % define only if not defined already % makes sure that this bit is compilable on its own but does not overwrite % other definitions \begin{proof} % of theorem Orientability of continuous line fields As \cref{thm:projection_as_a_covering_map} shows, \ver{$P ⫶ 𝕊M → 𝒬^{𝕊}M$}{$P_N ⫶ 𝕊^{N-1} → 𝒬^{𝕊'}ℝ^N$} is a covering map. Use $Q_{p_0}$ and a $n_0 ∈ 𝕊M$ with $P(n_0) = Q_{p_0}$ as the distinguished points. Connected manifolds are path-connected and locally path-connected.\ask{it's correct, do I need to proof it? Not hard but needs some lines.} % $Q_* ⸨ π_1(M, p_0) ⸩ ⊆ P_* ⸨ π_1(𝕊M, n_0) ⸩$ (for $n_0 ∈ 𝕊_{p_0}M$ such that $Q_{p_0} = P(n_0)$) % is equivalent to all generators of $π_1(M, p_0)$ being in $P_* ⸨ π_1(𝕊M, n_0) ⸩$. For the \enquote{if} part let $γ ∈ G$, \ie $γ ⫶[0,1] → M$ continuous with $γ(0) = γ(1) = p_0$. $Q$ is orientable along $γ$. That means that there exists $n ∈$\ver{$Γ_C(\rest{𝕊M}{γ([0,1])})$}{$C(γ([0,1]), 𝕊^{N-1})$} with $P(n) = Q$. If the orientation of $Q$ along this path has $n(γ(0)) = -n_0$, use $-n$ instead. Then $n ∘ γ$ is a continuous loop at $n_0$ in $𝕊M$ with $P_* [n ∘ γ] = [P ∘ n ∘ γ] = [Q ∘ γ] = Q_* [γ]$ Since $γ$ is an arbitrary generator of $π_1(M, p_0)$ and $Q_*$ and $P_*$ are group homomorphisms, we get that $Q_*(π_1(M, p_0)) ⊆ P_*(π_1(𝕊M, n_0))$. Hence there exists $n ⫶ C(M, 𝕊M)$ with $P(n) = Q$ by \cref{thm:lifting_of_continuous_maps}, \ie $Q$ is orientable \ver{and $n ∈ Γ_C(𝕊M)$}{}. For the \enquote{and only if} part restrict an orientation for $Q$ on the paths in $G$. \end{proof} % of theorem Orientability of continuous line fields \docEnd
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