@@ -186,6 +186,48 @@ Let $`Q ∈ W^{1,p}(Ω, 𝒬), 1 ≤ p ≤ ∞`$ be non-orientable. Then there e

- In chapter 3, orienting a line field is translated into lifting $`Q`$ from $`ℝP^2`$ to covering space $`𝕊^2`$.

- Classical algebraic topology tools only consider continous maps, but we have $`W^{1,p}(Ω)`$ maps and want to preserve this regularity

### Continous line fields

Chapter 3.1 deals with the lifting of continous line fields.

Similar to "normal" algebraic topology.

*Lemma 3*:

- $`Q : [t_1, t_2] → 𝒬`$ continous path can be lifted in two ways (depending on choice of start).

- If $`Q`$ stays close ($`≤ ε s`$ for $`0 < ε < \sqrt{2}`$) to some $`\overline{Q} ∈ 𝒬`$, then the lifted $`n`$ stays on the same half-circle as the start value ($`|n(t) - \overline m| ≤ ε`$)

*Proof notes for Lemma 3*: First calculation:

```math

| n ⊗ n - m ⊗ m |^2 = Σ_{i,j} | n_i n_j - m_i m_j |^2 = Σ_{i,j} ( n_i n_j - m_i m_j )^2

Define $`n`$ arbitrary (possibly non-continous) as $`Q(t) = s(n(t) ⊗ n(t) - \frac13\operatorname{Id})`$. Construct $`n^+`$ by starting at $`n(t_1)`$, then staying on the same side of the circle as $`n^+(t_1)`$ as long ("$`δ`$") as it is possible to prove. That makes it continous. Then start over at $`n^+(t_1 + δ)`$. This process is finite since a continous path into a bounded set on a compact domain is uniformly continous.

There are only two continous liftings because otherwise we would have a jump by 180° where it changes from $`n^+`$ to $`n^-`$.

Given that $`Q`$ stays close to some $`\overline{Q}`$, calculation shows that $`n^+ · m`$ can never be 0, so must be always $`>0`$.

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*Proposition 5*: $`Ω`$ simply connected, $`Q`$ continous. Then there exists a continous lifting $`n: P(n) = Q`$.

*Proof*: See algebraic topology: q4-q23 Proposition 1.33, currently p. 70

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*Theorem 1*: If $`Q: G := Ω \setminus \bigcup_{i = 1}^N \overline{ω_i} → 𝒬_2`$ is orientable at all $`∂ω_i`$, then $`Q`$ is orientable.

*Proof sketch*:

- For $`f:[0,1] → \overline{Ω}`$ define $`f^*`$ by replacing parts of $`f`$ in ω_i by (shorter) parts on $`∂ω_i`$. $`f^*`$ is continous (with proof in Lemma 4, but really?).

-*Lemma 5*:

1) $`\overline{Ω}`$ is path-connected and simply connected. Proof: Schoenflies theorem: Jordan curve creates homeomorphism to unit disk.

2) $`\overline{G}`$ is path-connected. Proof: see previous bullet point.

3) $`\overline{G}`$ is locally path-connected: for every point $`x ∈ \overline{G}`$ and every radius $`ε`$, there is a ball $`B_{δ}(x)`$, such that all points in $`B_{δ}(x) ∩ \overline{G}`$ are path-connected to each other with paths within $`B_{ε}(x) ∩ \overline{G}`$ is path-connected. Proof: Use homeomorphism from Schoenflies theorem, build small balls and a straight line in unitdisk.