[proofplan] We compute the transition function between the two oriented stereographic trivializations of $TS^2$ on the equator, using stereographic coordinates chosen to agree with the outward orientation. The overlap map is complex inversion, and the oriented clutching convention used here compares the southern tangent frame to the northern tangent frame along the equator oriented by increasing northern stereographic argument. Differentiating in that comparison direction gives, after parametrizing the equator by the northern coordinate $e^{i\theta}$ for $\theta\in\mathbb{R}$, rotation through angle $2\theta+\pi$. Removing the constant factor $-1$ by homotopy leaves the degree-two loop in $SO(2)$. [/proofplan]

[step:Compute the stereographic transition map on the overlap]Let $N:=(0,0,1)$ and $S:=(0,0,-1)$ denote the north and south poles of the unit sphere $S^2\subset\mathbb{R}^3$. Define the chart domains by \begin{align*} U_N := S^2 \setminus \{N\}. \end{align*} Define also \begin{align*} U_S := S^2 \setminus \{S\}. \end{align*} We regard $\mathbb{C}$ as $\mathbb{R}^2$ with its standard orientation, and we orient $S^2$ by the outward unit normal. Define the northern stereographic coordinate map $\varphi_N:U_N\to\mathbb{C}$ by \begin{align*} \varphi_N(X,Y,Z) = \frac{X-iY}{1-Z}. \end{align*} Define the southern stereographic coordinate map $\varphi_S:U_S\to\mathbb{C}$ by \begin{align*} \varphi_S(X,Y,Z) = \frac{X+iY}{1+Z}. \end{align*} These charts are oriented for the outward orientation: if $F_N:=\varphi_N^{-1}$ and $F_S:=\varphi_S^{-1}$ are written in real coordinates, direct differentiation gives $\partial_x F_N\times\partial_y F_N$ as a positive multiple of $F_N$ and $\partial_x F_S\times\partial_y F_S$ as a positive multiple of $F_S$. Let $z$ denote the coordinate on $\varphi_N(U_N)$ and $w$ denote the coordinate on $\varphi_S(U_S)$. The inverse map $\varphi_N^{-1}:\mathbb{C}\to S^2\setminus\{N\}$ is \begin{align*} \varphi_N^{-1}(z) = \left(\frac{2\operatorname{Re} z}{1+|z|^2},-\frac{2\operatorname{Im} z}{1+|z|^2},\frac{|z|^2-1}{1+|z|^2}\right). \end{align*} Therefore, for $z \in \mathbb{C}\setminus\{0\}$, substitution into $\varphi_S$ gives \begin{align*} (\varphi_S \circ \varphi_N^{-1})(z)=\frac{\frac{2\operatorname{Re}z}{1+|z|^2}-i\frac{2\operatorname{Im}z}{1+|z|^2}}{1+\frac{|z|^2-1}{1+|z|^2}}. \end{align*} Simplifying the numerator and denominator gives \begin{align*} (\varphi_S \circ \varphi_N^{-1})(z)=\frac{2\bar z}{1+|z|^2}\cdot \frac{1+|z|^2}{2|z|^2}=\frac{\bar z}{|z|^2}=\frac{1}{z}. \end{align*} Thus the oriented coordinate transition map is the holomorphic map $\tau:\mathbb{C}\setminus\{0\}\to \mathbb{C}\setminus\{0\}$ defined by \begin{align*} \tau(z)=z^{-1}. \end{align*}[/step]

[guided]We first restate the two stereographic coordinate maps used in the computation. Let $N:=(0,0,1)$ and $S:=(0,0,-1)$ be the poles of $S^2\subset\mathbb{R}^3$, and define $U_N:=S^2\setminus\{N\}$ and $U_S:=S^2\setminus\{S\}$. The northern chart is the map $\varphi_N:U_N\to\mathbb{C}$ given by \begin{align*} \varphi_N(X,Y,Z)=\frac{X-iY}{1-Z}. \end{align*} The southern chart is the map $\varphi_S:U_S\to\mathbb{C}$ given by \begin{align*} \varphi_S(X,Y,Z)=\frac{X+iY}{1+Z}. \end{align*} The signs in these two stereographic charts are chosen to match the outward orientation on $S^2$. In the northern chart, using $X-iY$ rather than $X+iY$ makes the ordered coordinate frame agree with the outward normal orientation near the south pole. In the southern chart, using $X+iY$ gives the same agreement near the north pole. Equivalently, if $F_N:=\varphi_N^{-1}$ and $F_S:=\varphi_S^{-1}$ are the inverse parametrizations, direct differentiation shows that $\partial_xF_N\times\partial_yF_N$ is a positive multiple of $F_N$ and $\partial_xF_S\times\partial_yF_S$ is a positive multiple of $F_S$. We compute the overlap map directly. If $z=\varphi_N(X,Y,Z)$, then the inverse stereographic formula gives the point $(X,Y,Z)\in S^2\setminus\{N\}$ as \begin{align*} (X,Y,Z)=\left(\frac{2\operatorname{Re} z}{1+|z|^2},-\frac{2\operatorname{Im} z}{1+|z|^2},\frac{|z|^2-1}{1+|z|^2}\right). \end{align*} Substituting this point into $\varphi_S$ gives \begin{align*} (\varphi_S \circ \varphi_N^{-1})(z)=\frac{\frac{2\operatorname{Re}z}{1+|z|^2}-i\frac{2\operatorname{Im}z}{1+|z|^2}}{1+\frac{|z|^2-1}{1+|z|^2}}. \end{align*} Simplifying the numerator and denominator gives \begin{align*} (\varphi_S \circ \varphi_N^{-1})(z)=\frac{2\bar z}{1+|z|^2}\cdot \frac{1+|z|^2}{2|z|^2}=\frac{\bar z}{|z|^2}=\frac{1}{z}. \end{align*} Thus, on the overlap $U_N\cap U_S$, the change from the northern complex coordinate $z$ to the southern complex coordinate $w$ is $w=1/z$. This explicit formula is the source of the clutching degree.[/guided]

[step:Differentiate the transition map to obtain the tangent transition]The tangent trivialization determined by an oriented chart identifies each tangent plane with $\mathbb{R}^2 \cong \mathbb{C}$ through the ordered coordinate frame. Hence the tangent transition from the northern trivialization to the southern trivialization is the real-[linear map](/page/Linear%20Map) represented by the complex derivative of $\tau$. Since \begin{align*} \tau(z)=z^{-1}, \end{align*} its derivative at $z\in \mathbb{C}\setminus\{0\}$ is complex multiplication by \begin{align*} \tau'(z)=-z^{-2}. \end{align*} Under the oriented clutching convention fixed in the statement, the clutching map assigns to each equatorial point the change-of-trivialization map from the southern oriented tangent frame to the northern oriented tangent frame. This is the reverse of the coordinate transition comparison, and since $\tau^{-1}=\tau$, it is multiplication by \begin{align*} (\tau^{-1})'(w)=-w^{-2}. \end{align*} On the equator we parametrize the clutching circle by the northern coordinate $z=e^{i\theta}$. Because $w=1/z=e^{-i\theta}$ on this circle, the southern-to-northern comparison is multiplication by \begin{align*} -w^{-2}=-e^{2i\theta}. \end{align*} Let $GL^+(2,\mathbb{R})$ denote the group of real $2\times2$ matrices with positive determinant. Thus the clutching loop $g:S^1\to GL^+(2,\mathbb{R})$ is \begin{align*} g(e^{i\theta})\colon v\mapsto -e^{2i\theta}v, \end{align*} where $v\in\mathbb{C}\cong\mathbb{R}^2$.[/step]

[guided]The tangent transition is obtained by differentiating the coordinate transition, because a coordinate chart trivializes the tangent bundle by sending the ordered coordinate basis in the chart to the standard ordered basis of $\mathbb{R}^2\cong\mathbb{C}$. From the previous step, the coordinate transition from the northern coordinate to the southern coordinate is the holomorphic map $\tau:\mathbb{C}\setminus\{0\}\to\mathbb{C}\setminus\{0\}$ given by \begin{align*} \tau(z)=z^{-1}. \end{align*} For $z\in\mathbb{C}\setminus\{0\}$, differentiating this holomorphic map gives the real-linear tangent map represented by complex multiplication by \begin{align*} \tau'(z)=-z^{-2}. \end{align*} The clutching convention in the theorem compares the southern tangent trivialization to the northern tangent trivialization, so we need the derivative in the reverse direction. Since $\tau^{-1}=\tau$, if $w$ is the southern coordinate then the reverse transition is multiplication by \begin{align*} (\tau^{-1})'(w)=-w^{-2}. \end{align*} Now orient the equatorial clutching circle by increasing northern stereographic argument and write the northern coordinate as $z=e^{i\theta}$ with $\theta\in\mathbb{R}$. On this circle the southern coordinate is \begin{align*} w=\frac{1}{z}=e^{-i\theta}. \end{align*} Therefore the southern-to-northern tangent comparison is multiplication by \begin{align*} -w^{-2}=-e^{2i\theta}. \end{align*} Thus, writing $GL^+(2,\mathbb{R})$ for the group of real $2\times2$ matrices with positive determinant, the clutching loop $g:S^1\to GL^+(2,\mathbb{R})$ is \begin{align*} g(e^{i\theta})\colon v\mapsto -e^{2i\theta}v, \end{align*} where $v\in\mathbb{C}\cong\mathbb{R}^2$.[/guided]

[step:Retract the transition loop to rotations and read off its winding number] For $\alpha\in\mathbb{R}$, let $R_\alpha\in SO(2)$ denote rotation of $\mathbb{R}^2\cong\mathbb{C}$ through angle $\alpha$, equivalently multiplication by $e^{i\alpha}$. We use the standard degree of a loop in $SO(2)\cong S^1$, namely the [winding number](/page/Winding%20Number) of its angle lift divided by $2\pi$. For every $\theta\in\mathbb{R}$, the map \begin{align*} v \mapsto -e^{2i\theta}v \end{align*} is already the rotation $R_{2\theta+\pi}$. Hence the loop already takes values in $SO(2)$, so no retraction from $GL^+(2,\mathbb{R})$ is needed. Define the rotation loop $h:S^1\to SO(2)$ by \begin{align*} h(e^{i\theta})=R_{2\theta+\pi}. \end{align*} The constant factor $-1$ contributes the additive constant $\pi$ to the angle. The homotopy $H:S^1\times[0,1]\to SO(2)$ defined by \begin{align*} H(e^{i\theta},t)=R_{2\theta+(1-t)\pi} \end{align*} is well-defined and deforms $h$ to the loop $e^{i\theta}\mapsto R_{2\theta}$. As $\theta$ increases from $0$ to $2\pi$, the angle $2\theta$ increases from $0$ to $4\pi$, so the loop winds twice around $SO(2)\cong S^1$. Hence the oriented clutching degree of $TS^2$ is $2$. [/step]