Let $(X,\tau), (Y,\sigma)$ be topological spaces with $|X|$ infinite and suppose $\varphi:X\to Y$ is a bijection such that for all $x\in X$ we have that $(X\setminus\{x\}) \cong (Y\setminus\{\varphi(x)\})$.


Does this imply that $(X,\tau) \cong (Y,\sigma)$?



Reconstructibility of topological spaces
The answer is ``No''. Let $X$ be the Cantor set and let $Y = X \setminus \{p\}$, where $p$ is any element of $X$. Let $\varphi \colon X \to Y$ be any bijection. Since the Cantor set with one point removed is homeomorphic to the Cantor set with two points removed, the given condition is satisfied. However, $X$ is compact whereas $Y$ is not. 


**Edit:** In response to the question by Loïc Teyssier, there are many ways to see that the assertion is true. One simple way is to note that if two points are removed from the Cantor set, the one-point compactification of the resulting space is compact, metrizable, zero-dimensional, and has no isolated points, and is, therefore, homeomorphic to the Cantor set. Hence, when the point at infinity is removed, the resulting space is homeomorphic to the Cantor set with a single point removed, but it is also the Cantor set with two points removed.


