Symmetric functions possess a common zero.

Question

So possess a common zero, means $\exists x \in S^k: f_i(x) = 0$, right?

Then I could not follow this brief proof - what is the corollary? Because the information in the proof is so little, I couldn't even guess which.

Thanks =)

Theorem. Any $k$ smooth functions $f_1,\ldots, f_k$ on $S^k$ that satisfy the symmetry condition $f_i(-x)=-f_i(x)$, $i=1,\ldots, k$, must possess a common zero.

Proof. If not, apply the corallary to the map $$ f(x)=(f_1(x),\ldots, f_k(x),0),$$ taking the $x_{k+1}$ axis for $l$.

Edit: the corollary turns out to be the theorem mentioned in this question.

Answer 1

Yes, with this theorem (why do they call it corollary, then...?), it works exactly like they say, by contradiction. If the $f_j$'s don't vanish simultaneously, then $f$ takes $S^k$ to $\mathbb{R}^{k+1}\setminus\{0\}$, and satisfies the symmetry condition. So by the corollary-theorem, it must intersect in particular the line $\{x_1=\ldots=x_k=0\}$. But this gives a simultaneous zero. Contradiction.