Fixed-point property is retract-hereditary

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.

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.