Hairy "disk" theorem

Question

Given a disk $D=\{x\in\mathbb{R}^2||x|\leqslant1\}$. There is a continuous tangent vector field $X:D\to\mathbb{R}^2$, which is always pointing towards outside of the disk on $\partial D$. To prove is there exist a zero of $X$ inside $D$.

The given solution applied Poincaré-Hopf theorem. I understand the main idea, but there is one point I don't understand.

Because Poincaré-Hopf theorem can't be used on $D$, so we set another vector field $Y:=-X$, and glue two copies of $D$ along $\partial D$ to identify an $S^2$. And then we get a resulted continuous vector field $Z$ on $S^2$. Here, I don't quite understand how to generate this $Z$? And two to glue them?

Generally, how to transform from hairy "disk" to hairy ball?

Answer 1

Let $D_{+}$ and $D_{-}$ be two copies of the closed unit disk, and put $r(x,y) = \sqrt{x^{2} + y^{2}}$. The mappings $$ f_{\pm}(x, y) = \left(\frac{x\sin(\pi r(x, y)/2)}{r(x, y)}, \frac{y\sin(\pi r(x, y)/2)}{r(x, y)}, \pm\cos(\pi r(x, y)/2)\right) $$ smoothly associate $D_{+}$ with the upper hemisphere and $D_{-}$ with the lower hemisphere. (The gymnastics with trig functions ensure the derivative is bounded at the boundary of the disk, cf. the graph parametrization.)

If $X$ is a vector field in $D_{+}$ that "points straight outward" in the sense that $X(x, y) = \lambda(x, y)(x, y)$ on the boundary of $D_{+}$ for some continuous, real-valued function $\lambda$, then

1. $(f_{+})_{*}X$ is a continuous vector field on the upper hemisphere;

2. The vector field $Y = -X$ induces a continuous vector field $(f_{-})_{*}(Y)$ on the lower hemisphere.

3. On the equator, $(f_{+})_{*}X = (f_{-})_{*}Y$.

The unit sphere may be viewed as the identification space $(D_{+} \sqcup D_{-})/\sim$ under the equivalence relation that identifies corresponding boundary points in $D_{\pm}$. Item 3. says there is a continuous vector field on the sphere given by $X$ (or really, its push-forward by $f_{+}$) in the upper hemisphere, and by $Y = -X$ in the lower hemisphere.