Let $A$ and $B$ be two points not on the real line $\mathbb{R}$. Consider the set $S=(\mathbb{R}\setminus\{0\})\cup\{A,B\}$.
For any two positive real numbers $c,d$ define $$I_A(-c,d)=]-c,0[\cup\{A\}\cup]0,d[.$$
Define a topology on $S$ as follows: On $\mathbb{R}\setminus\{0\}$ use the subspace topology inherited from $\mathbb{R}$, with open intervals as a basis. A basis of neighborhoods at $A$ is the set $\{I_A(-c,d)\;|\;c,d>0\};$. I must prove that the map $h\colon I_A(-c,d)\to (-c,d)$ defined by
\begin{split}
h(x)&=x\quad\text{for}\;x\in]-c,0[\;\cup\;]0,d[\\
h(A)&=0
\end{split}
is a homeomorphism.
Clearly $h$ is bijective and continuous.
Now, we must prove that $h^{-1}$ is continuous or equivalently, that $h$ is an open map.
My attempt.
Let $U$ be an open set of $S$, if $A\notin U$, that is $U$ is an open set of $\mathbb{R}\setminus\{0\}$, then $h(U)=U$ which is open.
If $A\in U$, then exists $c,d>0$ such that $(-c,0)\cup\{A\}\cup(0,d)\subseteq U$, in general $U=V\cup (-c,0)\cup \{A\}\cup (0,d)$, where $V$ is some open set in $\mathbb{R}\setminus\{0\}.$
Then $h(U)=h(V)\cup(-c,0)\cup(0,d)$ which is open in $\mathbb{R}\setminus\{0\}$.
Doing it this way is as if we had considered $h\colon S\to\mathbb{R}\setminus \{0\}$. I'm not sure if it's correct or not, could someone help me?
Best Answer
In the case $A\in U$, the equality $h(U)=h(V)\cup (-c,0)\cup(0,d)$ is not true, you have $h(U)=h(V)\cup (-c,d)$ which is open in $\Bbb R$. Like this you can conclude.
To give you a direction for a complete proof: You can prove that $h:S-\{B\}\to \Bbb R$ is a homeomorphism and it will follow that $h:I_A(-c,d)\to (-c,d)$ is a homeomorphism (because it's the restriction of a homeomorphism). As you said, $h$ is bijective and it is pretty clear from the definition that $h$ is continuous. To show that it is an open map, you can show that $h$ sends elements of a basis of the topology of $S-\{B\}$ to open sets. It will follow that $h$ is open because any open set $U$ can be writen $U=\cup_\alpha C_\alpha$ for $C_\alpha$ in the basis and $h(U)=h(\cup_\alpha C_\alpha)=\cup_\alpha h(C_\alpha)=open$.
Choose this basis to be the collection of open intervals that doesn't contain $0$ and the $I_A(-c,d)$. Now open intervals that doesn't contain $0$ are sent to themselves by $h$, so they are still open. Finally $h(I_A(-c,d))=(-c,d)$ which is open.