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On each line $l \subset \mathbb{R}^n, n \geqslant 2$ passing through $0$ we choose a point $a(l)$ such that $a(l)$ depends continuously on $l$. I have to prove that there exists a line with $a(l)=0$.

I show it using Borsuk-Ulam theorem. We can consider the set of lines passing through the origin in $\mathbb{R}^n$ as sphere $\mathbb{S}^{n-1}$ (I fix this ambiguity later). We assign to each point on a sphere a real number $\lambda$ (point on a line passing through the origin and $x$ is $\lambda x$ for some $\lambda$). So we have a continuous function $$\varphi:\mathbb{S}^{n-1} \to \mathbb{R}$$ such that $\varphi(-x) = -\varphi (x)$, so it is odd. I take the embedding of $\mathbb{R}$ into $\mathbb{R}^{n-1}$ given by $x \longmapsto (x,0,\dots,0)$ which is continuous and this gives me a continuous odd function from $\mathbb{S}^{n-1} \to \mathbb{R}^{n-1}$. Finally, Borsuk-Ulam claims that there should be a point on a sphere with $\varphi(x)=\varphi(-x)$ and clearly, by oddity $\varphi(x)=0$ for some $x$. Am I correct? Thanks in advance.

Nicholas S
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You have to make precise the meaning of "$a(l)$ depends continuously on $l$". The only reasonable explanation is that you take the set $\mathbb{RP}^{n-1}$ of all lines through $0$ (i.e. of all one-dimensional subspaces) and give it the identification topology induced by the surjection $p : \mathbb{R}^n \backslash \{ 0 \} \to \mathbb{RP}^{n-1}, p(x) = $ line through $0$ and $x$. This space is the real projective space of dimension $n-1$.

Then your claim is that for each continuous $a : \mathbb{RP}^{n-1} \to \mathbb{R}^n$ such that $a(l) \in l$ for all $l$ there exists $l$ such that $a(l) = 0$.

Assume there exists a map $a$ such that $0 \ne a(l) \in l$ for all $l$. Define $a' : \mathbb{RP}^{n-1} \to S^{n-1}, a'(x) = a(x)/\lVert a(x) \rVert$. Then $p \mid_{S^{n-1}} \circ \phantom {.} a' = id$. This is impossible, but I do not see that it is a consequence of the Borsuk-Ulam theorem. One can easily show that $p' : p \mid_{S^{n-1}} : S^{n-1} \to \mathbb{RP}^{n-1}$ is a two-sheeted covering (between connected spaces). But $a'$ is section of $p'$ which implies that $p'$ is one-sheeted (see If a covering map has a section, is it a $1$-fold cover?). This contradiction shows that our assumption was wrong.

Paul Frost
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