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The theorem states that (call it theorem A)

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

The proof is based on a theorem (call it theorem B)

If $g: S^k\to R^{k+1}-\{0\}$ is symmetric about the origin ($g(-x) = -g(x)$), then $g$ intersects every line through $0$ at least once.

by making $f(x) = (f_1(x),f_2(x),\dots, f_k(x), 0)$ and pick line as the $x_{k+1}$ axis.

My naive question is that theorem B requires that the codomian of $g$ does not contain the origin; while in the proof of theorem A we invoke theorem B and find a zero point of $f$. I'm confused if it vilotates the condition?

Another question is that in theorem $A$ if we make $f_1(x)=x$ be the identity map on $S^1$, why would it have a zero point? Many thanks.

Edit: Thanks for @Mariano Suárez-Álvarez 's reply in the comments below. For the first question, I realize it's in the way we perform the proof. We should show theorem A by contradiction, that is assuming the function $f(x) = (f_1(x),f_2(x),\dots, f_k(x), 0)$ has no zero point. Then it is valid to invoke B. By picking the line as the $x_{k+1}$ axis, the assumption $f(x)$ has no zero point contracts theorem B. For the second question, the functions $f_i$ is actually $f_i:S^k\to R$, then my example is not valid.

naive101
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1 Answers1

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To close the question, I copy the edits from above.

For the first question, I realize it's in the way we perform the proof. We should show theorem A by contradiction, that is assuming the function $f(x) = (f_1(x),f_2(x),\dots, f_k(x), 0)$ has no zero point. Then it is valid to invoke B. By picking the line as the $x_{k+1}$ axis, the assumption $f(x)$ has no zero point contracts theorem B.

For the second question, the functions $f_i$ is actually $f_i:S^k\to R$, then my example is not valid.

naive101
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