Fixed-point property

A mathematical object X has the fixed-point property if every suitably well-behaved mapping from X to itself has a fixed point. It is a special case of the fixed morphism property. The term is most commonly used to describe topological spaces on which every continuous mapping has a fixed point. But another use is in order theory, where a partially ordered set P is said to have the fixed point property if every increasing function on P has a fixed point.

1 Definition
2 Examples
- 2.1 The closed interval
- 2.2 The closed disc
3 Topology
4 References

Definition

Let A be an object in the concrete category C. Then A has the fixed-point property if every morphism (i.e., every function) $f:A\to A$ has a fixed point.

The most common usage is when C=Top is the category of topological spaces. Then a topological space X has the fixed-point property if every continuous map $f:X\to X$ has a fixed point.

Examples

The closed interval

The closed interval [0,1] has the fixed point property: Let f:[0,1] → [0,1] be a continuous mapping. If f(0) = 0 or f(1) = 1, then our mapping has a fixed point at 0 or 1. If not, then f(0) > 0 and f(1) − 1 < 0. Thus the function g(x) = f(x) − x is a continuous real valued function which is positive at x = 0 and negative at x = 1. By the intermediate value theorem, there is some point x₀ with g(x₀) = 0, which is to say that f(x₀) − x₀ = 0, and so x₀ is a fixed point.

The open interval does not have the fixed-point property. The mapping f(x) = x² has no fixed point on the interval (0,1).

The closed disc

The closed interval is a special case of the closed disc, which in any finite dimension has the fixed-point property by the Brouwer fixed-point theorem.

Topology

A retract A of a space X with the fixed-point property also has the fixed-point property. This is because if $r:X\to A$ is a retraction and $f:A\to A$ is any continuous function, then the composition $i\circ f\circ r:X\to X$ (where $i:A\to X$ is inclusion) has a fixed point. That is, there is $x\in A$ such that $f\circ r(x)=x$ . Since $x\in A$ we have that $r(x)=x$ and therefore $f(x)=x.$

A topological space has the fixed-point property if and only if its identity map is universal.

A product of spaces with the fixed-point property in general fails to have the fixed-point property even if one of the spaces is the closed real interval.

The FPP is a topological invariant, i.e. is preserved by any homeomorphism. The FPP is also preserved by any retraction.

According to Brouwer fixed point theorem every compact and convex subset of an euclidean space has the FPP. Compactness alone does not imply the FPP and convexity is not even a topological property so it makes sense to ask how to topologically characterize the FPP. In 1932 Borsuk asked whether compactness together with contractibility could be a necessary and sufficient condition for the FPP to hold. The problem was open for 20 years until the conjecture was disproved by Kinoshita who found an example of a compact contractible space without the FPP.^[1]

References

^ Kinoshita, S. On Some Contractible Continua without Fixed Point Property. Fund. Math. 40 (1953), 96–98

Samuel Eilenberg, Norman Steenrod (1952). Foundations of Algebraic Topology. Princeton University Press.
Schröder, Bernd (2002). Ordered Sets. Birkhäuser Boston.