Fixed-Point Property is Topological

Theorem

Let $T \sim T'$ be homeomorphic topological spaces.

Suppose every continuous $f : T \to T$ has a fixed point.

Then, every continuous $g : T' \to T'$ also has a fixed point.

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Proof

Let $\phi : T \to T'$ be a homeomorphism.

Let $g : T' \to T'$ be an arbitrary continuous mapping.

Consider $f_g : T \to T$ defined as:

$\map {f_g} x = \map {\phi^{-1}} {\map g {\map \phi x}}$

By definition of homeomorphism and Composite of Continuous Mappings is Continuous:

$f_g$ is continuous

Therefore, by hypothesis, there is a fixed point $x_0 \in T$ of $f_g$.

We then have:

\(\ds \map {f_g} {x_0}\)

\(=\)

\(\ds x_0\)

Definition of Fixed Point

\(\ds \leadsto \ \ \)

\(\ds \map \phi {\map {f_g} {x_0} }\)

\(=\)

\(\ds \map \phi {x_0}\)

\(\ds \leadsto \ \ \)

\(\ds \map \phi {\map {\phi^{-1} } {\map g {\map \phi {x_0} } } }\)

\(=\)

\(\ds \map \phi {x_0}\)

Definition of $f_g$

\(\ds \leadsto \ \ \)

\(\ds \map g {\map \phi {x_0} }\)

\(=\)

\(\ds \map \phi {x_0}\)

Definition of Inverse Mapping

Therefore, $\map \phi {x_0}$ is a fixed point of $g$.

As $g$ was arbitrary, the result holds.

$\blacksquare$