$X,Y$ not homeomorphic if $X\backslash x$ is disconnected for all $x\in X$ but $Y\backslash y$ is connected for some $y\in Y$

Question

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 ... ?

Answer 1

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.

Answer 2

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?

Answer 3

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?