Commit 842cb814 authored by Felix Hilsky's avatar Felix Hilsky
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add version of alternative proof

parent c6c4ed4e
%! TEX program = lualatex
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\providecommand*{\ver}[2]{#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{$P ⫶ 𝕊M → 𝒬^{𝕊}M$}{$P_N ⫶ 𝕊^{N-1} → 𝒬^{𝕊'}^N$} 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
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