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Does a homeomorphism between two topological spaces need to map clopen sets to clopen sets?

For example consider the Baker‘s map $$ (x,y)\mapsto (2x mod 1, 1/2(y+\lfloor 2x\rfloor)) $$ $B\colon X\to X$ on $X=[0,1]\times [0,1]$ with the standard topology, as well as the space shift-map $\sigma\colon Y\to Y$ with $Y=\{0,1\}^{\mathbb{Z}}$ equipped with the product topology.

Then the map $\varphi\colon Y\to X$ defined by

$$ (\ldots,s_{-2},s_{-1},s_0,s_1,s_2,\ldots)\mapsto (x,y)=(\sum_{i=0}^\infty \frac{s_i}{2^{i+1}},\sum_{i=1}^{\infty}\frac{s{-i}}{2^i}) $$

is a homeomorphism.

Now what is the image of a cylinder set, i.e. a set $$ C_{a_1,…,a_k}^{n_1,…,n_k}=\{(s_i)\in Y: s_{n_j}=a_j: j=1,…,k \} $$

Such cylinder sets are clopen but its Image cannot be clopen since X does not have clopen sets in the standard topology (despite of X and the emptyset).

But since $\varphi^{-1}$ is also continuous, the Image of a cylinder set needs to be clopen!

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    $\begingroup$ Why do you think that $\varphi$ is a homeomorphism? It cannot be exactly for the reason you pointed out. (Or simpler because $Y$ is disconnected while $X$ is connected) $\endgroup$
    – Alessandro Codenotti
    Mar 3, 2022 at 13:08
  • $\begingroup$ But the Bakers map and the Shift are topologically conjugated, aren‘t they? $\endgroup$
    – Salamo
    Mar 3, 2022 at 13:11
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    $\begingroup$ Some authors use topologically conjugate to mean that there exist a continuous injection $\varphi\colon Y\to X$ with $\varphi\circ\sigma=B\circ\varphi$, without requiring $\varphi$ to be a homeomorphism, maybe this is where the confusion is coming from $\endgroup$
    – Alessandro Codenotti
    Mar 3, 2022 at 13:20
  • $\begingroup$ But the map $\varphi$ is bijective... where does it fail to be a homeomorphism? $\endgroup$
    – Salamo
    Mar 3, 2022 at 15:09
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    $\begingroup$ Continuous bijections are not necessarily homeomorphisms. Indeed, the inverse may fail to be continuous, which is precisely what is needed to preserve clopen sets. $\endgroup$
    – Sambo
    Mar 3, 2022 at 15:23

1 Answer 1

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Yes to the question in the first line: a homeomorphism is both an open and a closed map, so it preserves clopen sets.

So if your $\phi$ does not, it cannot be a homeomorphism.

$Y$ is (totally) disconnected and $X$ is connected so there cannot be a homeomorphism between them.

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  • $\begingroup$ I see. I would like to understand which properties of a homeomorphism the function $\varphi$ has (and which not). As far as I see, $\varphi$ is continuous and bijective. Then, its inverse $\varphi^{-1}$ cannot be continuous. Is there a way to see this without using that, otherwise, one would have a homeomorphism which is impossible due to the conectedness/ discommectedness? $\endgroup$
    – Salamo
    Mar 3, 2022 at 20:08
  • $\begingroup$ Moreover, if there is a continuous bijective function (that is no homeomorphism) between topological Spaces such that the diagram commutes : how is that called, semi-conjugation? But this Name is reserved for continuous surjections… $\endgroup$
    – Salamo
    Mar 3, 2022 at 20:10

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