Fixed-point property is retract-hereditary: Difference between revisions

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* Compose with the retraction to get a self-map of the whole space
* Compose with the retraction to get a self-map of the whole space
* Find a fixed point, and observe that it must be a fixed point of the original self-map
* Find a fixed point, and observe that it must be a fixed point of the original self-map
===Proof details===
'''Given''': A topological space <math>X</math> satisfying the fixed-point property, a retraction <math>r:X \to A</math> where <math>A \subset X</math> and <math>r(a) = a</math> for all <math>a \in A</math>
'''To prove''': <math>A</math> satisfies the fixed-point property
'''Proof''': Let <math>i</math> denote the inclusion of <math>A</math> in <math>X</math>.
Consider any continuous map <math>f:A \to A</math>. We need to show that <math>f</math> has a fixed point in <math>A</math>. Consider the composition <math>g = i \circ f \circ r</math>. This is a map from <math>X</math> to <math>X</math> that first retracts to <math>A</math>, then applies <math>f</math>, and then views the resulting point of <math>A</math> as a point in <math>X</math>. <math>g</math> is a composite of continuous maps, so <math>g</math> is continuous. Since <math>X</math> has the fixed-point property, there exists <math>x \in X</math> such that <math>g(x) = x</math>.
But by construction, <math>g(x)</math> is actually inside <math>A</math>, so in fact <math>x \in A</math>. But if <math>x \in A</math>, <math>r(x) = x</math>, so we conclude that <math>x = g(x) = f(r(x)) = f(x)</math>. Thus, <math>x \in A</math> is a fixed point of <math>f</math>, completing the proof.

Revision as of 20:00, 20 July 2008

This article gives the statement, and possibly proof, of a topological space property satisfying a topological space metaproperty
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Statement

Property-theoretic statement

The property of topological spaces called the fixed-point property is a retract-hereditary property of topological spaces.

Verbal statement

Any retract of a topological space having the fixed-point property, also has the fixed-point property.

Definitions used

Fixed-point property

Retract

Subspace topology

Proof

Proof outline

  • Consider a self-map of the retract
  • Compose with the retraction to get a self-map of the whole space
  • Find a fixed point, and observe that it must be a fixed point of the original self-map

Proof details

Given: A topological space X satisfying the fixed-point property, a retraction r:XA where AX and r(a)=a for all aA

To prove: A satisfies the fixed-point property

Proof: Let i denote the inclusion of A in X.

Consider any continuous map f:AA. We need to show that f has a fixed point in A. Consider the composition g=ifr. This is a map from X to X that first retracts to A, then applies f, and then views the resulting point of A as a point in X. g is a composite of continuous maps, so g is continuous. Since X has the fixed-point property, there exists xX such that g(x)=x.

But by construction, g(x) is actually inside A, so in fact xA. But if xA, r(x)=x, so we conclude that x=g(x)=f(r(x))=f(x). Thus, xA is a fixed point of f, completing the proof.