I have seen here on stack exchange (in the comments) a proof along the lines of:

$X\backslash x$ is disconnected for all $x\in X$ but $Y\backslash y$ is connected for some $y\in Y$. Therefore $X$ and $Y$ are not homeomorphic.

Explicitly, how does this show the nonexistence of a homeomorphism between $X$ and $Y$?

A homeomorphism is a bijective continuous function with continuous inverse.
A continuous function's inverse maps open sets to open sets.
A connected space is the union of disjoint open sets.

Prior attempts: Write out the definitions, then ... ?

  • 3
    $\begingroup$ Hi Caleb -- to get the best response, it is usually best to add context to your questions. What do you know about homeomorphisms? What do you know about disconnectedness? Have you tried to prove this yourself, and if so, what happened? $\endgroup$ Apr 5, 2019 at 2:19
  • $\begingroup$ Your title says the exact opposite of what your body says. The title claims that the conclusion is they are homeomorphic, but you are saying you are trying to prove they are not homeomorphic. $\endgroup$ Apr 5, 2019 at 3:24
  • $\begingroup$ $X/x$ is disconnected for all $x$ implies that $X$ is simply-connected, on other hand if $Y/y$ is disconnected for some $y$ implies $Y$ is multiple-connected. Connectedness is an topological invariant and therefore $X$ and $Y$ are not related by an homeomorphism $\endgroup$
    – user473029
    Apr 5, 2019 at 3:44
  • $\begingroup$ chat.stackexchange.com/transcript/message/60749431#60749431 $\endgroup$ Mar 28 at 9:11

3 Answers 3


Here explicitly:

Let $(X,\tau_X),(Y,\tau_Y)$ be the topological spaces $X,Y$ with their considered topologies $\tau_X,\tau_Y$.

Assume there is a homeomorphism $h: (X,\tau_X)\rightarrow(Y,\tau_Y)$, while $Y\setminus \{y\}$ is connected but $X\setminus \{x\}$ is disconneted with $x = h^{-1}(y)$.

$X\setminus \{x\}$ disconnected means you can split

  • $X\setminus \{x\} = A\cup B$ where $A\neq \emptyset$ and $B\neq \emptyset$ and $A,B \in \tau_{X\setminus \{x\}}$, which means they are open and closed wrt. $\tau_{X\setminus \{x\}}$ - the topology on $X\setminus \{x\}$ induced by $\tau_X$.
  • Since $h$ is a homeomorphism, it follows that $Y\setminus \{y\} = h(A) \cup h(B)$ and $h(A)$ and $h(B)$ are nonempty and they are both open and closed in $\tau_{Y\setminus \{y\}}$ - the topology on $Y\setminus \{y\}$ induced by $\tau_Y$.

Hence, $Y\setminus \{y\}$ is now disconnected, which is a contradiction to the assumption that it is not.


Hint: Suppose that there exists a homeomorphism $f : X \to Y$ and let $$x := f^{-1}(y).$$ Can you show that $X \setminus \{x\}$ must be homeomorphic to $Y \setminus \{y\}$? What can be said about the continuous image of a connected space?


There are two relevant facts:

  1. Continuous functions whose domains are restricted are continuous.
  2. Continuous functions take connected sets to connected sets.

Can you use these facts to prove that the circle is not homeomorphic to the line?


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