1

I have the following problem:

Problem: Show that the following statements are equivalent:

  1. If $f:\mathbb{S}^{n}\longrightarrow\mathbb{S}^{m}$ is a continuous function such that $f(-x)=-f(x)$, then $n\leq m$.

  2. If $f:\mathbb{S}^{n}\longrightarrow\mathbb{R}^{n}$ is a continuous function, then there exists $x\in\mathbb{S}^{n}$ such that $f(x)=f(-x)$. (Borsuk-Ulam Theorem)

  3. If $\mathbb{S}^{n}$ is covered by $n+1$ closed sets $A_{1},...,A_{n+1}$ then at least one of the $A_{i}$ contains an antipodal pair of points. (Lusternik-Schnirelman Theorem).

  4. If $f:\mathbb{S}^{n}\longrightarrow\mathbb{S}^{n}$ is a continuous function such that $f(-x)=-f(x)$, then $f$ has odd degree.

  5. If $f:\mathbb{S}^{n}\longrightarrow\mathbb{R}^{n}$ is a continuous function such that $f(-x)=-f(x)$, then there exists $x_{0}\in \mathbb{S}^{n}$ such that $f(x_{0})=0$.

The implications $(1)\Rightarrow (2)\Rightarrow (3)$, $(1)\Rightarrow (5)$ and $(2)\Leftrightarrow (5)$ are not complicated, but I don't know how to finish testing the remaining implications. Although from the form of the statements, I think that $(3)$ implies $(1)$ would be perhaps the most viable option.

Any hint will help me. Thanks!

Fernando Mauricio Rivera Vega
  • 445
  • 4
    All these statement are true, and all true statements are logically equivalent. What you mean is to give "simple proofs" for these implications instead of proving all statements individually. – Paul Frost Nov 09 '21 at 00:57
  • @PaulFrost: sorry, What do you mean by "all true statements are logically equivalent"? By "all" you are referring to the OP 5 statements or something else? – C.F.G Nov 09 '21 at 05:41
  • Why do you think $4$ is equivalent to B-U theorem? – one potato two potato Nov 09 '21 at 05:46
  • 1
    If $m < n$ then $S^m$ injects naturally into $\mathbb{R}^n$. You can use this to show that (5) implies (1), and thus (2),(1),(5) are equivalent. Note that (4) implies that $f$ is surjective, which implies, in turn, (1), since $S^m$ embeds naturally into $S^n$ for $m < n$.

    Then you can show that (3) implies that the image of every odd continuous map $S^n \rightarrow S^n$ contains a vector with equal coefficients with the correct $A_i$. By the same reasoning as above (1) follows. And so you know that (1),(2),(3),(5) are equivalent implied by (4). I’m not sure yet about the final part.

     – Aphelli Nov 09 '21 at 09:06
  • 1
    @C.F.G I refer to all true statements, e.g. $0. \quad 1 + 1 = 2$. Then the implications $0. \Rightarrow 1., 2., 3.,4., 5.$ are logically true (simply because 1. - 5. are true), but there is certainly no simple proof of 1. - 5. based only the fact $1 + 1 = 2$. – Paul Frost Nov 09 '21 at 10:01
  • @Mindlack You should give an official answer, even if it still has a gap. – Paul Frost Nov 09 '21 at 10:03
  • @PaulFrost: You mean one can prove any theorem just by $1+1=2$? [that implication is true logically but the correctness of statement is not because of 0. it is?] – C.F.G Nov 09 '21 at 10:25
  • @C.F.G As I said "there is certainly no simple proof of 1. - 5. based only the fact $1+1=2$". But on the level of formal logic each true statement $A$ implies each other true statement $B$. So if we know that $A$ is true but do not know that $B$ is true, then we need a proof for $B$. Sometimes this is very simple (e.g. the truth of $1 + 1 = 2$ implies the truth of $1 + 1 + 1 +1 = 2 + 2$), in other cases it may be hard work. For example the truth of $1 +1 = 2$ does not help us to prove 1. - 5. What I wanted to say is this: To prove the equivalence of 1. - 5. we can give an individual proof – Paul Frost Nov 09 '21 at 11:48
  • for each of them. This will show that they are logically equivalent. But this is not very efficient, it is better to find simple arguments showing that e.g. 1. $\Rightarrow$ 2. instead of proving 2. ab ovo. – Paul Frost Nov 09 '21 at 11:49
  • @PaulFrost I think it is clear that a proof of each of the statements in the list is being sought as an implication of some other problem in the same list. If I am wrong, excuse me, from now on I will try to be very specific when writing so that they do not solve a problem for me with the fact that $1+1=2$. Thanks for your help. – Fernando Mauricio Rivera Vega Nov 09 '21 at 16:42
  • What is your definition of degree? Based on $f_* : H_n(S^n) \to H_n(S^n)$? – Kritiker der Elche Nov 09 '21 at 23:49

1 Answers1

2

This answer is incomplete, because I can’t see how to easily deduce (4) from the other statements – it looks to me like a significant strengthening of the theorem, but I may easily be looking at it from the wrong angle. Anyway, let’s see all the other implications:

$(4) \Rightarrow (1)$: by contradiction. If $f: S^n \rightarrow S^m$ is continuous and odd with $m<n$, its composition with the natural embedding $S^m \rightarrow S^n$ is still continuous and odd and nonsurjective, hence of degree zero, a contradiction.

$(1) \Rightarrow (2)$: by contradiction. If $f$ does not take the same value on two antipodal points, let $g(x)=\frac{f(x)-f(-x)}{|f(x)-f(-x)|} \in S^{n-1}$, then $g: S^n \rightarrow S^{n-1}$ is odd continuous, a contradiction.

$(2) \Rightarrow (3)$: consider $f(x)=(d(x,A_i))_{1 \leq i \leq n}$.

$(3) \Rightarrow (1)$: by contradiction, we can assume $m<n$. Let $g$ be the composition of $f$ with the inclusion $S^m \subset S^{n}$. Then $g$ is odd and $S^n \rightarrow S^n$.

For $0 \leq i \leq n$, let $A_i$ be the set of $x \in S^n$ such that the $i$-th coordinate of $g(x)$ is not less than any other one. Then the $A_i$ are closed subsets covering $S^n$ so one of them contains two antipodal points: it means that for some $x \in S^n$ and some $0 \leq i \leq n$, the $i$-th coordinate of $g(x)$ is the largest and the smallest, so that $g(x)$ has equal coordinates and is thus not in the image of $S^m$ in $S^n$: we get our desired contradiction.

$(2) \Rightarrow (5)$: if $f: S^n \rightarrow \mathbb{R}^n$ is odd, then there is (by (2)) some $x \in S^n$ such that $f(x)=f(-x)$. As $f(x)=-f(-x)$, $f(x)=0$.

$(5) \Rightarrow (2)$: let $f: S^n \rightarrow \mathbb{R}^n$, then $x \longmapsto f(x)-f(-x)$ is odd so vanishes at some $x \in S^n$, then $f(x)=f(-x)$.

Aphelli
  • 34,439
  • Very nice (+1). Note in $(4) \Rightarrow (1)$ you only used the weaker "$(4')$ If $f:\mathbb{S}^{n}\longrightarrow\mathbb{S}^{n}$ is a continuous function such that $f(-x)=-f(x)$, then $f$ has non-zero degree." So $(4')$ is one more property equivalent to the others. – Paul Frost Nov 10 '21 at 09:34
  • It may help that an odd map $f: S^n \to S^n$ induces a map $f' : \mathbb RP^n \to \mathbb RP^n$. See https://math.stackexchange.com/q/1784016 and https://planetmath.org/proofofborsukulamtheorem, but that seems to be not the type of proof desired here. It does not use 1. - 3. or 5. but is based on completely different methods. – Paul Frost Nov 10 '21 at 09:44
  • @Paul Frost: that’s a good point, but $(1) \Rightarrow (4’)$ isn’t entirely straightforward. The proof I found uses (1) for $S^{n+1}$: if $f:S^n \rightarrow S^n$ is continuous odd and with degree $0$, it is homotopic to a point (this in itself is already nontrivial). So we can “fill” the sphere, making it a continuous map $H: B_{n+1} \rightarrow S^n$ whose restriction to the boundary is odd. Define $g: S^{n+1} \rightarrow S^n$ by $g(x,u)=H(x)$ if $u \geq 0$ and $-H(-x)$ if $u \leq 0$ (where $x \in B_{n+1}, u \in \mathbb{R}$ and $|x|^2+u^2=1$). Then $g$ is continuous odd, contradicting (1). – Aphelli Nov 10 '21 at 09:47
  • Regarding your other comment on the proof of (4), they seem to be proofs of (4) “from first principles”, not really “deductions” from close-sounding statements. – Aphelli Nov 10 '21 at 09:51
  • Right, these proofs are in a sense "ab ovo" - though perhaps not as drastic as in my comments to the question ;-) But that does not exclude that there is a simple deduction from the other statements, although I doubt it. – Paul Frost Nov 10 '21 at 10:00
  • See also exercise 4 in https://www.dpmms.cam.ac.uk/study/III/Algebraic%20Topology/2018-2019/AT(III)2018-Sheet2.pdf. – Paul Frost Nov 10 '21 at 10:12