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For non-arbitary spaces we can discuss for such case, like how many components are there or other properties. But is it true for any space? It seems if we have a homeomorphism $f$ from $S$ to $S' = S - \{p\}$, $f(p) = q$, but since a space is homeomorphic to itself, there is some $g(r)=q$. Then there is no inverse if $f$ and $g$ coincide. However they don't have to and maybe $f$ is somehow the homeomorphism since I can't deduce more information.

If it is not true, a counter-example will be super helpful! Thank you.

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    $\begingroup$ Even two different puncturings of a space needn’t be homeomorphic; it may matter which point is removed. Consider removing a point from a figure “X”. The result might have 1, 2, or 4 components, depending on which point is removed. $\endgroup$ – MPW Dec 24 '19 at 21:32
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Any infinite discrete space is homeomorphic to itself minus any point. For example, the map $n\mapsto (n+1)$ is a homeomorphism $\mathbb{N}\to \mathbb{N}\setminus\{0\}$ (where $\mathbb{N}$ has the discrete topology).

The same example works if you give $\mathbb{N}$ the trivial topology or the Alexandrov topology (where open sets are upwards-closed sets).

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  • $\begingroup$ Oh sure, discrete spaces always work. Thanks! $\endgroup$ – MonkeyKing Dec 24 '19 at 20:34
  • $\begingroup$ @MonkeyKing Not finite ones! $\endgroup$ – Alex Kruckman Dec 24 '19 at 20:36
  • $\begingroup$ Sure, I mean, I should always think about discrete space as my first example to check something... $\endgroup$ – MonkeyKing Dec 24 '19 at 20:37
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Another example: $\mathbb C \setminus \mathbb Z$. This is connected and homeomorphic to the punctured version of itself.

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$\mathbb{Q}$ is homeomorphic to its punctured version (all countable metric spaces without isolated points are). Same for the irrationals $\mathbb{P}$.

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