I was attempting to do an exercise on my topology book. It states the following:

Let $f:\mathbb{R} \mapsto \mathbb{R}$ be a homeomorphism such that $\exists p \in \mathbb{N}: f^p =$ identity. Prove that $f$ has a fixed point.

I think I have a solution but I never really use the fact that $f$ is a homeomorphism. That is:

Let $g(x)=f(x)-x$. Then $g$ is a continuous function. Now, let's say that $f(x)\geq x \ \forall x \in \mathbb{R}$. Then $f^p(x) \geq f(x) \geq x = f^p(x) \forall x$, which implies $f(x)=x \forall x$.

In the same way, if $f(x)\leq x \forall x$ we have $f^p(x) \leq f(x) \leq x = f^p(x) \implies f(x) = x$.

So, if $f(x)$ is not the identity, then $\exists x_1, x_2 \in \mathbb{R}: f(x_1) < x_1, f(x_2) > x_2$, and because $g$ is continuous and $g(x_1)<0, g(x_2)>0$ then there's a point $\bar{x}: g(\bar{x})=0 \implies f(\bar{x})=\bar{x}$, so it is a fixed point.

I've not actually used the fact that $f$ is a homeomorphism, I only used the fact that it is continuous and has finite order, so is my proof wrong?

  • 8
    $\begingroup$ I think your proof is fine. Note that being continuous and having finite order already implies that $f$ is a homeomorphism, since $f^{p-1}$ is a left- and right-inverse. $\endgroup$ – PhoemueX Jul 27 at 16:24
  • $\begingroup$ That's a fair point indeed $\endgroup$ – DottorMaelstrom Jul 27 at 16:29

Your proof is correct, but in my opinion it could be formulated more transparently. In fact, you consider three cases.

Case 1: $\forall x \in \mathbb{R} : f(x) \ge x$.

Case 2. $\forall x \in \mathbb{R} : f(x) \le x$.

Case 3: Neither 1 nor 2 is satisfied. Then $\exists x_1, x_2 \in \mathbb{R} : f(x_1) < x_1, f(x_2) > x_2$. Now introduce $g$.


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