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@WlodekKuperberg (perhaps) and I (for sure) were exposed to this kind of] examples by Karol Borsuk, or possibly Karol Borsuk simply had an example like the one I will present below:

\begin{equation} D\ :=\ \{z\in\mathbb C: |z|\le 1\}\ \subseteq\ \mathbb C \ \end{equation} \begin{equation} A\,\ :=\,\ D\times\{0\}\ \cup\ \{1\ \ \ i\ \ -\!1\ \ -\!i\}\times [-1;0]\,\ \subseteq\,\ \mathbb C\times\mathbb R \end{equation} \begin{equation} X\,\ :=\,\ A\,\ \cup\,\ \{-1\ \ 1\}\times [0;1]\,\ \subseteq\,\ \mathbb C\times\mathbb R \end{equation} \begin{equation} Y\,\ :=\,\ A\,\ \cup\,\ \{i\ \ \ 1\}\times [0;1]\,\ \subseteq\,\ \mathbb C\times\mathbb R \end{equation}

Then $\ X\ $ and $\ Y\ $ are not homeomorphic while $\ X\times I\ $ and $\ Y\times I\ $ are.

REMARK One may check Karol Borsuk's series of publications about the uniqueness of topological decomposition into Cartesian products, and a paper by Hanna Patkowska about the uniqueness of the decomposition of ANRs into 1-dimensional ANRs.

A kind request (I'd greatly appreciate): Wlodek Kuperberg, please add a picture to my analytic description; let the pictures of $\ X\ $ and $\ Y\ $ be embedded into $\ \mathbb C;\ $ I mean homeomorphic copies of $\ X\ $ and $\ Y$.

ACKNOWLEDGEMENT I am grateful to Wlodek Kuperberg for providing such a very nice graphics (so cute and psychologically loaded; it's the first graphics illustration in my MO posts)). *** Włodek, congratulation on your another NICE answer (Gauss said, a few but ripe).

@WlodekKuperberg (perhaps) and I (for sure) were exposed to this kind of examples by Karol Borsuk, or possibly Karol Borsuk simply had an example like the one I will present below:

\begin{equation} D\ :=\ \{z\in\mathbb C: |z|\le 1\}\ \subseteq\ \mathbb C \ \end{equation} \begin{equation} A\,\ :=\,\ D\times\{0\}\ \cup\ \{1\ \ \ i\ \ -\!1\ \ -\!i\}\times [-1;0]\,\ \subseteq\,\ \mathbb C\times\mathbb R \end{equation} \begin{equation} X\,\ :=\,\ A\,\ \cup\,\ \{-1\ \ 1\}\times [0;1]\,\ \subseteq\,\ \mathbb C\times\mathbb R \end{equation} \begin{equation} Y\,\ :=\,\ A\,\ \cup\,\ \{i\ \ \ 1\}\times [0;1]\,\ \subseteq\,\ \mathbb C\times\mathbb R \end{equation}

Then $\ X\ $ and $\ Y\ $ are not homeomorphic while $\ X\times I\ $ and $\ Y\times I\ $ are.

REMARK One may check Karol Borsuk's series of publications about the uniqueness of topological decomposition into Cartesian products, and a paper by Hanna Patkowska about the uniqueness of the decomposition of ANRs into 1-dimensional ANRs.

A kind request (I'd greatly appreciate): Wlodek Kuperberg, please add a picture to my analytic description; let the pictures of $\ X\ $ and $\ Y\ $ be embedded into $\ \mathbb C;\ $ I mean homeomorphic copies of $\ X\ $ and $\ Y$.

ACKNOWLEDGEMENT I am grateful to Wlodek Kuperberg for providing such a very nice graphics (so cute and psychologically loaded; it's the first graphics illustration in my MO posts)). *** Włodek, congratulation on your another NICE answer (Gauss said, a few but ripe).

@WlodekKuperberg (perhaps) and I (for sure) were exposed to this kind of] examples by Karol Borsuk, or possibly Karol Borsuk simply had an example like the one I will present below:

\begin{equation} D\ :=\ \{z\in\mathbb C: |z|\le 1\}\ \subseteq\ \mathbb C \ \end{equation} \begin{equation} A\,\ :=\,\ D\times\{0\}\ \cup\ \{1\ \ \ i\ \ -\!1\ \ -\!i\}\times [-1;0]\,\ \subseteq\,\ \mathbb C\times\mathbb R \end{equation} \begin{equation} X\,\ :=\,\ A\,\ \cup\,\ \{-1\ \ 1\}\times [0;1]\,\ \subseteq\,\ \mathbb C\times\mathbb R \end{equation} \begin{equation} Y\,\ :=\,\ A\,\ \cup\,\ \{i\ \ \ 1\}\times [0;1]\,\ \subseteq\,\ \mathbb C\times\mathbb R \end{equation}

Then $\ X\ $ and $\ Y\ $ are not homeomorphic while $\ X\times I\ $ and $\ Y\times I\ $ are.

REMARK One may check Karol Borsuk's series of publications about the uniqueness of topological decomposition into Cartesian products, and a paper by Hanna Patkowska about the uniqueness of the decomposition of ANRs into 1-dimensional ANRs.

A kind request (I'd greatly appreciate): Wlodek Kuperberg, please add a picture to my analytic description; let the pictures of $\ X\ $ and $\ Y\ $ be embedded into $\ \mathbb C;\ $ I mean homeomorphic copies of $\ X\ $ and $\ Y$.

@WlodekKuperberg (perhaps) and I (for sure) were exposed to this kind of] examples by Karol Borsuk, or possibly Karol Borsuk simply had an example like the one I will present below:

\begin{equation} D\ :=\ \{z\in\mathbb C: |z|\le 1\}\ \subseteq\ \mathbb C \ \end{equation} \begin{equation} A\,\ :=\,\ D\times\{0\}\ \cup\ \{1\ \ \ i\ \ -\!1\ \ -\!i\}\times [-1;0]\,\ \subseteq\,\ \mathbb C\times\mathbb R \end{equation} \begin{equation} X\,\ :=\,\ A\,\ \cup\,\ \{-1\ \ 1\}\times [0;1]\,\ \subseteq\,\ \mathbb C\times\mathbb R \end{equation} \begin{equation} Y\,\ :=\,\ A\,\ \cup\,\ \{i\ \ \ 1\}\times [0;1]\,\ \subseteq\,\ \mathbb C\times\mathbb R \end{equation}

Then $\ X\ $ and $\ Y\ $ are not homeomorphic while $\ X\times I\ $ and $\ Y\times I\ $ are.

REMARK One may check Karol Borsuk's series of publications about the uniqueness of topological decomposition into Cartesian products, and a paper by Hanna Patkowska about the uniqueness of the decomposition of ANRs into 1-dimensional ANRs.

A kind request (I'd greatly appreciate): Wlodek Kuperberg, please add a picture to my analytic description; let the pictures of $\ X\ $ and $\ Y\ $ be embedded into $\ \mathbb C;\ $ I mean homeomorphic copies of $\ X\ $ and $\ Y$.

ACKNOWLEDGEMENT I am grateful to Wlodek Kuperberg for providing such a very nice graphics (so cute and psychologically loaded; it's the first graphics illustration in my MO posts)). *** Włodek, congratulation on your another NICE answer (Gauss said, a few but ripe).

>@WlodekKuperberg (perhaps) and I (for sure) were exposed to this kind of examples by Karol Borsuk, or possibly Karol Borsuk simply had an example like the one I will present below:

\begin{equation} D\ :=\ \{z\in\mathbb C: |z|\le 1\}\ \subseteq\ \mathbb C \ \end{equation} \begin{equation} A\,\ :=\,\ D\times\{0\}\ \cup\ \{1\ \ \ i\ \ -\!1\ \ -\!i\}\times [-1;0]\,\ \subseteq\,\ \mathbb C\times\mathbb R \end{equation} \begin{equation} X\,\ :=\,\ A\,\ \cup\,\ \{-1\ \ 1\}\times [0;1]\,\ \subseteq\,\ \mathbb C\times\mathbb R \end{equation} \begin{equation} Y\,\ :=\,\ A\,\ \cup\,\ \{i\ \ \ 1\}\times [0;1]\,\ \subseteq\,\ \mathbb C\times\mathbb R \end{equation}

Then $\ X\ $ and $\ Y\ $ are not homeomorphic while $\ X\times I\ $ and $\ Y\times I\ $ are.

REMARK One may check Karol Borsuk's series of publications about the uniqueness of topological decomposition into Cartesian products, and a paper by Hanna Patkowska about the uniqueness of the decomposition of ANRs into 1-dimensional ANRs.

A kind request (I'd greatly appreciate): Wlodek Kuperberg, please add a picture to my analytic description; let the pictures of $\ X\ $ and $\ Y\ $ be embedded into $\ \mathbb C;\ $ I mean homeomorphic copies of $\ X\ $ and $\ Y$.

@WlodekKuperberg (perhaps) and I (for sure) were exposed to this kind of] examples by Karol Borsuk, or possibly Karol Borsuk simply had an example like the one I will present below:

\begin{equation} D\ :=\ \{z\in\mathbb C: |z|\le 1\}\ \subseteq\ \mathbb C \ \end{equation} \begin{equation} A\,\ :=\,\ D\times\{0\}\ \cup\ \{1\ \ \ i\ \ -\!1\ \ -\!i\}\times [-1;0]\,\ \subseteq\,\ \mathbb C\times\mathbb R \end{equation} \begin{equation} X\,\ :=\,\ A\,\ \cup\,\ \{-1\ \ 1\}\times [0;1]\,\ \subseteq\,\ \mathbb C\times\mathbb R \end{equation} \begin{equation} Y\,\ :=\,\ A\,\ \cup\,\ \{i\ \ \ 1\}\times [0;1]\,\ \subseteq\,\ \mathbb C\times\mathbb R \end{equation}

Then $\ X\ $ and $\ Y\ $ are not homeomorphic while $\ X\times I\ $ and $\ Y\times I\ $ are.

REMARK One may check Karol Borsuk's series of publications about the uniqueness of topological decomposition into Cartesian products, and a paper by Hanna Patkowska about the uniqueness of the decomposition of ANRs into 1-dimensional ANRs.

A kind request (I'd greatly appreciate): Wlodek Kuperberg, please add a picture to my analytic description; let the pictures of $\ X\ $ and $\ Y\ $ be embedded into $\ \mathbb C;\ $ I mean homeomorphic copies of $\ X\ $ and $\ Y$.