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I've been reading this article that proposes an inductive proof of Brouwer's fixed point theorem just using "basic" topology, avoiding the usage of homotopy groups as the usual proof goes.

To do so we assume that given $n \in \mathbb{N}$ any continuous function from $C = [0,1]^n$ to $C$ has a fixed point and we prove that any continuous function $f$ from $C\times[0,1]$ to $C \times [0,1]$ has a fixed point (the base case for $n = 1$ is already proven by the intermediate value theorem). We write $f = (f_1,\dots,f_{n+1})$ where each $f_i:C\times [0,1] \rightarrow [0,1]$ is a continous function. Next it defines $\phi = (f_1,\dots,f_n)$ which is clearly a continuous function and we write $\phi_t: C \rightarrow C$ as the function $\phi_t(x) = \phi(x,t)$, which clearly is continuous. Next by induction hypothesis we note that each $\phi_t$ has a fixed point, with this in mind we define $$X = \{(x,t) \in C \times [0,1]: \phi_t(x) = x\} = (\phi-id)^{-1}(\{0\})$$ And because this set is the preimage of a closed set under a continuous function, it's closed and nonempty because by induction each $\phi_t$ has at least one fixed point.

However here's where my problem with the proof comes because next it takes $\pi_2: C\times [0,1] \rightarrow [0,1]$ as the projection onto the last coordinate. We note that $\pi_2(X) = [0,1]$ as there's at least one fixed point for each $\phi_t$, but the problem that because clearly $\pi_2$ is an open map the proof on the article asumes that $\pi_2|_X$ is also an open map. However, restrictions of open maps are not always open, so is the article wrong or what am I missing?

I've tried to prove that this map is in fact open, but I can't seem to do so, and in fact it looks that it might not be open which would prove that the article is wrong, so how do you prove that in fact this restriction is open? Or what would be a counterexample of a function $f$ such that the set we defined as $X$ is such that the restriction of $\pi_2$ isn't open?

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    $\begingroup$ Please write in paragraphs. Thanks $\endgroup$ Oct 26, 2023 at 7:44
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    $\begingroup$ @freakish Your comments should be an official answer. $\endgroup$
    – Paul Frost
    Oct 26, 2023 at 8:23
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    $\begingroup$ @Fshrike This proof is based on Sperner's Lemma. $\endgroup$
    – Paul Frost
    Oct 26, 2023 at 10:10
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    $\begingroup$ If such a really elementary proof were correct, it would have been a sensation! $\endgroup$
    – Ulli
    Oct 26, 2023 at 11:16
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    $\begingroup$ And good job on being skeptical. But note that even if a proposed proof asserts only true statements, it does not imply that it is correct. It is not the case that you can debunk any wrong proof by proving the negation of some statement it asserts! If a proof is wrong, it is wrong simply because it does not obey logical reasoning, and for no other reason at all. $\endgroup$
    – user21820
    Oct 28, 2023 at 13:20

2 Answers 2

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The claim that $\pi_2$ restricted to $X$ is open is unfortunately false. Consider

$$f:[0,1]^2\to[0,1]^2$$ $$f(x,t)=(tx,t)$$

Then $\phi(x,t)=tx$ and $\phi_t(x)=tx$. Therefore $\phi_t(x)=x$ if and only if $x=0$ or $t=1$. In other words

$$X=\big(\{0\}\times[0,1]\big)\cup\big([0,1]\times\{1\}\big)$$

You can verify very easily that neither of the projections restricted to $X$ is open (some of its open subsets are mapped onto a point).


After reading the article this looks like a crucial, unrecoverable mistake in the proof. Or at least I don't see how this can be fixed. I might be wrong though. Or maybe there is no elementary proof after all...

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    $\begingroup$ Providing an example in which $\pi_2 \mid_ X$ is not open shows that the proof does not work and cannot repaired. Perhaps there is a completely different elementary proof (though I doubt it), but the present approach is inadequate. $\endgroup$
    – Paul Frost
    Oct 26, 2023 at 8:56
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The answer by freakish has already identified the concrete flaw in the claimed proof. But there also is a fundamental obstacle to any kind of "elementary inductive proof", namely that Brouwer's Fixed Point theorem in dimension 1 behaves very different than in higher dimensions.

If we consider a computable function $f : [0,1] \to [0,1]$, it will always have a computable fixed point (either it has an isolated fixed point, which then is computable by bisection, or it has an entire interval full of fixed points, which will contain computable numbers as well). If we could prove Brouwer's Fixed Point theorem by iterating the intermediate value theorem somehow, we would expect that this proof also shows that every computable function $f : [0,1]^n \to [0,1]^n$ has a computable fixed point.

However, Orevkov has constructed an example of a computable function $f : [0,1]^2 \to [0,1]^2$ having no computable fixed point at all.

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  • $\begingroup$ Can you link a place where I can find this example of a computable function? $\endgroup$
    – H4z3
    Oct 27, 2023 at 0:46
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    $\begingroup$ @H4z3 The original paper is quite inaccessible. There is a nice writeup by Petrus Potgieter here: arxiv.org/abs/0804.3199 You may also be interested in this paper: arxiv.org/abs/1206.4809 $\endgroup$
    – Arno
    Oct 27, 2023 at 6:37

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