# Fixed-point property is retract-hereditary

This article gives the statement, and possibly proof, of a topological space property (i.e., fixed-point property) satisfying a topological space metaproperty (i.e., retract-hereditary property of topological spaces)
View all topological space metaproperty satisfactions | View all topological space metaproperty dissatisfactions
Get more facts about fixed-point property |Get facts that use property satisfaction of fixed-point property | Get facts that use property satisfaction of fixed-point property|Get more facts about retract-hereditary property of topological spaces

## Statement

### Verbal statement

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

## 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 $r:X \to A$ where $A \subset X$ and $r(a) = a$ for all $a \in A$

To prove: $A$ satisfies the fixed-point property

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

Consider any continuous map $f:A \to A$. We need to show that $f$ has a fixed point in $A$. Consider the composition $g = i \circ f \circ r$. 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 $x \in X$ such that $g(x) = x$.

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