Regarding the winding number My main question is about part B, but I would also be grateful if you can tell me what you think about part A.

Define a smooth vector field
$X$ on $S^1$ as follows:
$X(x,y)=(-y,x)$. For a smooth map $f:S^1\to M$ we define $f^{'}(t)=d_tf(X_t)$ for $t\in S^1$.
Assume that there's given a smooth map
$\gamma:S^1\to R^2\setminus \{0\}$.  Define
$f:S^1\to S^1$ by
$f(t)=\frac{\gamma(t)}{\| \gamma(t)\|}$.
Recall, the winding number of $\gamma$ is defined as: $n(\gamma)=deg(f)$ where on both circles we put the standard orientation.


A. If $u\in R^2 \setminus \{0\}$, define  $R_u=\{au: a>0\}$. Show that there exists a vector $u\in R^2\setminus \{0\}$ such that for all $t\in \gamma^{-1}(R_u)$ the derivative $\gamma^{'}(t)$ is not parallel to $u$, show that for this $u$ the set $\gamma^{-1}(R_u)$ is finite.


B. If $u$ as in part A, for $t\in \gamma^{-1}(R_u)$  define $\epsilon(t)\in \{\pm 1\}$ to be the sign of the ordered basis $(u,\gamma^{'}(t))$. Prove that:

$n(\gamma)=\sum_{t\in \gamma^{-1}(R_u)} \epsilon(t)$.
My try:
Part A:
Method 1: I tried to use a theorem (1) that states:
If $M, N$ are smooth n-manifolds , $f:M\to N$ ia a smooth proper map, $y\in N$ is a regulr value of $f$ then $f^{-1}(y)$ is finite.
Regarding the question, $f$ is smooth since $\gamma$ is smooth and $\|\gamma\|$ is smooth (except in $0$ which does not belong to $S^1$). $f$ is also proper, by  a theorem: if $X,Y$ are topological spaces hausdorff and compact and $f:X\to Y$ is contiuos then for every compact $K\subset Y$ , $f^{-1}(K)$ is compact.
In our case $f$ is smooth so continuos and $S^1$ is compact and hausdorff therefore $f$ is proper. So using theorem (1) we get that there is a regular value $v\in S^1$ such that $f^{-1}(v)$ is finite.In particular $y\in R^2\setminus \{0\}$. Consider $R_y=\{ay: a>0\}$  and assume by contradiction that there is $t\in \gamma^{-1}(R_y)$ such that $\gamma'(t)$ is parallel to $y$ .
Since $f$ is the projection of $\gamma$ on $S^1$ so $d_t(f)=0$ and so $f$ is not surjective and this contradicts that $y$ is a regular value of $f$. Thus for this $y$, for all $t\in \gamma^{-1}(R_y)$ : $\gamma'(t)$ is not parallel to $y$.
In addition, $f^{-1}(y)$ is finite if
$t\in \gamma^{-1}(R_y)$ iff $\gamma(t) \in R_y$ iff $\frac{\gamma(t)}{\|\gamma(t)\|}=y$ iff $t\in f^{-1}(y)$
So $\gamma^{-1}(R_y)$ is finite.
Method 2:
Notice that $f$ is the composition of $\gamma$ with the radial quotient $q: R^2 \setminus \{0\} \to S^1$ ( $q$ is the map $R^2\setminus \{0\} \to S^1$ defined by $v\to \frac{v}{\|v\|}$ )..
This quotient just projects out the radial component of any vector (what does radial component mean in this context?), so the condition that $\gamma'(t)$ is not parallel to $u$ is equivalent to the condition that $f'(t)$ is nonzero, i.e. $f$ is a submersion at $t$. Also notice that $R_u$ is $q^{-1}(v)$ where $v := q(u) = \frac{u}{\|u\|}$, so the condition that $\gamma'(t)$ is not parallel to $u$ for all $t$ in $\gamma^{-1}(R_u) = f^{-1}(v)$ is equivalent to the condition that $v$ is a regular value of f. By Sard's Theorem, this holds for almost all $u$. For such a $u$, the usual preimage theorem says that $f^{-1}(v) = \gamma^{-1}(R_u)$ is a 0-manifold. It must be compact because $S^1$ is compact, so it's finite.
Part B:
Definition: if $M, N$ are smooth oriented n-manifilds without boundary and $f:M\to N$ a smooth map. For a regular value $y\in N$ of $f$ and under the condition $|f^{-1}(y)|<\infty$, define $deg_y(f)=\sum_{x\in f^{-1}(y)} sgn (d_xf)$  where $d_xf:T_xM\to T_yN$ an isomorphism ($T_xM, T_yN$ are oriented).
We have that $deg(f)=deg_v(f)$ by a theorem.
Translating as in Part A, the expression given on the RHS is equal to
$\sum_{t \in f^{-1}(v)} sgn (d_t f)$.
Since $v$ is a regular value for $f$, it suffices to show that $sgn(v, \gamma'(t)) = sgn(d_t f)$ for all $t \in f^{-1}(v)$. By subtracting off an appropriate multiple of $v$ from $\gamma'(t)$, you get $d_q(\gamma'(t)) = f'(t)$, so $(v, \gamma'(t))$ has the same orientation as $(v, f'(t))$. This orientation is $+1$ if $f'(t)$ is a positive multiple of $X_t$, and $-1$ if $f'(t)$ is a negative multiple of $X_t$, and these are equivalent to saying that $d_t f$ preserves or reverses orientation, respectively.
I did not understand how this works, what does it mean to subtract off an appropriate multiple of $v$ from $γ′(t)$?, and do how we get then that $d_q(\gamma'(t)) = f'(t)$?
Any clarification would be welcomed.
 A: Let $\gamma=(\gamma_1,\gamma_2)$. Define
$\Gamma: S^1 \times \mathbb{R}^+ \rightarrow \mathbb{R}^2\setminus \{0\}$ by
$(t,a)\rightarrow (a \gamma_1(t),a\gamma_2(t))$. Then
$D\Gamma = \begin{pmatrix} a\gamma_1'(t) & \gamma_1(t) \\ a\gamma_2'(t) & \gamma_2(t) \end{pmatrix}$ and
$\text{det}(D\Gamma) = a(\gamma_1'(t)\gamma_2(t) - \gamma_2'(t)\gamma_1(t))$.
As $a>0$,  $\gamma'(t)$ is not parallel to $\gamma(t)$ iff
$\text{det}(D\Gamma) \neq 0$
iff $u = \Gamma(t,a)$ is a regular value.
By Sard, $\exists\,u$ a regular value. If $u \notin \text{Image}(\Gamma)$ then $\gamma^{-1}(R_u)=\emptyset$ and  both (A) and (B) are trivially true (geometrically, $0$ is not contained in a bounded component of
$\mathbb{R}^2\setminus \text{Image}(\gamma)$ and hence $n(\gamma) = n(0,\gamma) = 0$).
So let $u=\Gamma(t,a)$ be a regular value. Since $S^1$ is compact, Image($\gamma$) is bounded and the distance of $\gamma$ from $0$ is also positive; ie, $\text{Image}(\gamma)$ is contained in an annulus $\{0 < a_1 < \lVert(x,y)\rVert < a_2  < \infty \}$. Then $\gamma^{-1}(R_u) \subset Y = S^1\times [a_1,a_2]$ and by restricting $\Gamma$ to $Y$ we get a proper map and hence $\gamma^{-1}(R_u)$ is finite. Therefore (A).
To deduce (B), let $g:S^1\rightarrow S^1\times \mathbb{R}^+$ by $t\rightarrow (t,1/\lVert \gamma(t) \rVert)$.  Then $f = \Gamma \circ g$ and $df = D\Gamma \circ dg$. Note $dg(\partial_t) = (1, *)$ does not reverse orientation and hence $sgn(df)=sgn(D\Gamma)$. Now $(u_1,u_2) = (a\gamma_1(t),a\gamma_2(t))$ and $a >0$, so $sgn(u,\gamma') = sgn(\gamma,\gamma')=sgn(df)$. Hence (B).
