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This is Rotman "Intro to Algebraic Topology" Exercise 0.4:

If $X$ is a topological space homeomorphic to $D^n$, then every continuous $f:X\to X$ has a fixed point.

How can this be shown?

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  • $\begingroup$ Did Rotman prove that it is true for $X = D^n$? $\endgroup$
    – Paul Frost
    Commented May 2, 2023 at 22:23
  • $\begingroup$ @PaulFrost Yes, it is given as Theorem 0.3 of his book, attributed to Brouwer. $\endgroup$
    – pyridoxal_trigeminus
    Commented May 2, 2023 at 22:36

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Let $f:X\to X$ be a continuous function. Since $X \cong D^n$, we have a homeomorphism (i.e., a continuous bijection with continuous inverse) $h:X \to D^n$. Consider the diagram $$ D^n \xrightarrow{h^{-1}} X \xrightarrow{f} X \xrightarrow h D^n, $$ and observe that the composition $g := h \circ f \circ h^{-1}$ is continuous as the composition of continuous functions. Therefore by Brouwer's Fixed Point Theorem, there exists $x\in D^n$ such that $g(x)=x$. As a consequence, $f(h^{-1}(x))= h^{-1}(x)$, so $f$ also has a fixed point.

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    $\begingroup$ You need to say homeomorphism where you have continuous bijection. $\endgroup$
    – Ted Shifrin
    Commented May 2, 2023 at 21:10
  • $\begingroup$ @TedShifrin Fixed now, cheers $\endgroup$
    – pyridoxal_trigeminus
    Commented May 2, 2023 at 21:13

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