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If $f:X \to X$ (codomain and domain have the same topology) is a continuous bijection and every point has finite orbit, is $f^{-1}$ continuous?

Note that the orbit being finite and $f$ being a bijection means for all $x$ means for all $x$ there is an $n>0$ such that $f^n(x)=x$.

I asked myself this question while answering another question on this site and ended up not getting anywhere. I suspect it should come from very basic facts or is false in general. I am leaning towards false at the moment but have not been able to construct a counter example.

One attempt I tried was to see if considering $X_F$, the set $X$ final topology with respect to $f,f^{-1}$ had the same topology, and I believe I got $f:X_F \to X$ was continuous but, could never get the same for $f^{-1}$.

This question is similar, without the finite orbit restriction.

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    If the orbits are uniformly bounded in size (as the point varies) then $f^{-1}$ is a power of $f$ (meaning by composition) and hence continuous. So a counter example would require the orbit sizes to be unbounded. – Seth Mar 14 '15 at 23:19
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    @Seth Yes, and I think that might help put some restriction on what a counter example could look like, a topology where you wouldn't be able to glue all the $f^{-1}$ restricted to the bounded orbits together and arrive at a continuous function. –  Mar 14 '15 at 23:28
  • Just for the education here (in particular I like Jim Belk's answer) are examples of continuous bijections between spaces with the same topology where the inverse is not continuous. I don't believe any of those example have finite orbits though. –  Mar 15 '15 at 00:05
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    I am curious as to why the question was downvoted. I am open to suggestions to improve the question. –  Mar 15 '15 at 02:49

1 Answers1

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This seems to be a counterexample. Take $X = \mathbb{N}^2 \cup \{\infty\}$ with the metric where $\rho((m,n), \infty) = 1/n$ and $\rho((m,n), (p, q)) = 1/n + 1/q$ when $(m,n) \ne (p,q)$. Define $f(\infty) = \infty$ and $$ f((m,n)) = \cases{ (m, n-1) & if $2 \le n \le m$, \\ (m, m) & if $n=1$, \\ (m, n) & otherwise. } $$ For every $m \in \mathbb{N}$ the set $\{ (m, 1), \dots, (m, m) \}$ is an finite orbit and any point not in such a set is fixed. Since $(\mathbb{N}^2, \rho)$ is discrete, we only need to consider continuity at $\infty$. It is easy to verify that $\rho(f(m,n), \infty) \le 2\rho((m,n), \infty)$, so $f$ is continuous. On the other hand, every neighbourhood of $\infty$ contains a point $(m,m)$ for some value of $m$ and $\rho(f^{-1}(m,m), \infty) = 1$, hence $f^{-1}$ is not continuous.

For a slightly more natural counterexample, you could construct a vector space automorphism of $\mathbb{R}^\infty$ in a similar way, for example such that $$\begin{eqnarray} Te_1 &=& 2e_2, Te_2 = \frac{e_1}{2}, \\ Te_3 &=& 2e_4, Te_4 = 2e_5, Te_5 = \frac{e_3}{4} \\ \text{etc.}&& \end{eqnarray} $$ Clearly $T$ is a bounded linear operator w.r.t. the Euclidean norm but its inverse is not bounded. To see that it has finite orbits, note that the orbits of basis vectors are finite, say $T^{k_i} e_i = e_i$, and it follows for $x = \sum_{i=1}^n c_ie_i$ that $T^K x = x$, where $K = \prod_{i=1}^n k_i$.

Niels J. Diepeveen
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