Does the Hairy Ball theorem imply the Borsuk-Ulam for even dimensions? Consider the Hairy Ball theorem, stating that a continuous vector field over $R^{2n}$ must have a point where the tangential component vanishes,
and the Borsuk-Ulam theorem, stating that any continuous function $f: S^k \to \mathbb{R}^k$ must send two antipodal points to the same image.
I was wondering if, for $k = 2n$, the Borsuk-Ulam theorem follows from the Hairy Ball theorem.
The idea would be to take a continuous function $f: S^{2n} \to \mathbb{R}^{2n}$ and define $g: S^{2n} \to \mathbb{R}^{2n}$ as $g(P) = f(P) - f(-P)$. If we think of $g$ as a tangential vector field on $S^{2n}$, it follows that $\exists P \in S^{2n}: g(P) = f(P) - f(-P) = 0$ and the Borsuk-Ulam theorem (for even dimensions) follows.
I am not sure if I can "think of $g$ as a tangential vector field on $S^{2n}$", otherwise I believe the proof is fine.
Any thoughts on this?
 A: WOW! I've learnt the hairy ball theorem several weeks ago and I've just been watching 3blue1brown's video on the Borsuk-Ulam theorem https://www.youtube.com/watch?v=yuVqxCSsE7c&t=82s and the idea just hit me! I googled and found this post! Now here're my naive thoughts on that:
We can cover the sphere $S^{2n}$ with patches $U_k:= S^{2n}-(0,\dots,0,1)$ whose chart map is the stereographic projection, and $ V_k:= S^{2n}\cap B^3_{\frac{1}{k}}(1,0,\dots,0) $ whose chart map is the projection to $B^2_{\frac{1}{k}}(0,\dots,0)$ . The charts with them give two continuous frame bundles $\{ \frac{\partial  }{\partial x^i}\},\{\frac{\partial  }{\partial y^i} \}$ over $U_k,V_k$ respectively.  Since $S^{2n}$ is a compact manifold, the open cover admits a partition of unity $\rho_{U_k}, \rho_{V_k}$.
Now write the value of $g$ as $g(p)= g^1(p) e^1 + \cdots + g^{2n}(p) e^{2n}$ and  define linear isomorphisms: $ \phi_x : e^i \mapsto \frac{\partial  }{\partial x^i}$ and $ \phi_y : e^i \mapsto \frac{\partial  }{\partial y^i}$ and define the tangent bundles $\phi_x \circ g (p)= g^1(p) \frac{\partial  }{\partial x^1} + \cdots + g^{2n}(p) \frac{\partial  }{\partial x^{2n}}$ over $ U_k$ and $\phi_y \circ g (p)= g^1(p) \frac{\partial  }{\partial y^1} + \cdots + g^{2n}(p) \frac{\partial  }{\partial y^{2n}}$ over $ V_k$. Since the coefficients and the frames are continuous, so are these tangent vector fields over open subsets.
Using the partition of unity, we obtain a global continuous tangent vector field: $ \rho_{U_k} \phi_x \circ g+ \rho_{V_k} \phi_y \circ g $. By the hairy ball theorem, this field vanishes at $p_k$ for some $p_k \in S^{2n} $.
If $p_k \in U_k-V_k  $, then since $\phi_x$ is a linear isomorphism, the vector (since $supp \rho_{ V_k}$ is disjoint from $p_k$) $\rho_{U_k} \phi_x \circ g(p_k) =  \phi_x \circ g(p_0) =0 \iff g(p_k) =0$.
If $p_k \in V_k$, then denote $V_{k+1}:= B_{\frac{1}{k+1}}(1,0,\dots,0)$ and repeat the process above thus find $p_{k+1}$ s.t. the continuous tangent vector field: $ \rho_{U_{k+1}} \phi_x \circ g+ \rho_{V_{k+1}} \phi_y \circ g $ vanishes at it. Again, if $p_{k+1} \in U_{k+1}-V_{k+1}$, the process terminates; If not , continue in this fashion...
Therefore we get a sequence of points $\{p_{k}\}$ tending to $(1,0,\dots,0)$ s.t. $$(\rho_{U_k} \phi_x \circ g+ \rho_{V_k} \phi_y \circ g) (p_k)=0.  $$ Now let $k \to \infty$, then $$ \phi_y \circ g (1,0,\dots,0)=0 \iff g(1,0,\dots,0)=0. $$
Thanks for Jason pointing out my gaps. I am not satisfied with the modification. My intuition tells me to make one of the open subset in the cover to be smaller and smaller to "push the point to the support of only one bump function". I don't think the argument is rigorous enough. Maybe when I learn deeper I might know a better one to approach this problem. Any ideas in the comment will be appreciated.
