Brouwer's fixed point theorem for permutations

Question

Brouwer's fixed point theorem: Let f:[a,b]->[a,b]. There exists some x in [a,b] such that f(x)=x.

The above theorem deals with functions going from a set to the same set. These functions are, by definition, permutations. Despite this being the case, one can think of permutations in which none of the points land on themselves. For example, the permutation (123) for a set with three elements. This permutation maps every element to its neighbor element. As such, one could argue that Brouwer's fixed point theorem does not hold for this function, despite the fact that it met all of the criteria.

To summarize my question, how do permutations fit into Brouwer's fixed point theorem, despite them apparently disproving the theorem?

Answer 1

The 1-dimensional BFPT states:

Suppose that $f: [a,b] \to [a,b]$. If $f$ is continuous, then $f$ has a fixed point.

This requires continuity. You could ask how it generalizes, and it does, but in a very particular way.

BFPT does not require that $f$ be bijective. It is easy to manufacture (even non-constant) continuous maps $f: [a,b] \to [a,b]$ that are not bijective but for which it is easy to see that the conclusion of BFPT holds.

You want to apply this to permutations of a finite set $\{1, 2, \ldots, n\}$. This is misguided for two reasons: (a) BFPT is not about bijections and (b) it is about continuous functions on compact intervals in the Euclidean line. There is nothing about the finite set $\{1, 2, \ldots, n\}$ that resembles the structure of $[a,b]$ at all, and there is no way to make sense of continuity on $\{1, 2, \ldots, n\}$ in any reasonable way that resembles anything Euclidean.

In short, permutations do not fit into BFPT and they do not disprove it.

Answer 2

As pointed out by Randall, Brouwer's fixed point theorem does not deal with bijections on a set $X$, but with continuous functions $f : X \to X$ on a topological space $X$.

On each set $X$ with more than one element you find a function (even a bijection) without a fixed point. So you see that topology and continuity play an essential role.

However, you can regard any set $X$ as a topogical space by giving it the discrete topology or the trivial topology. In both cases all functions $f : X \to X$ will be continuous. Your example shows that there exist spaces $X$ and continuous functions on $X$ which do not have a fixed point.

A space $X$ is said to have the fixed point property if each continuous $f : X \to X$ has a fixed point. See for example https://planetmath.org/fixedpointproperty. This is a very special property, and most spaces do not have it. Examples are all closed balls in any Euclidean space $\mathbb{R}^n$. To prove that a space has the fixed point property frequently requires to apply methods of algebraic topology.