The following proof is from Basic Topology by Armstrong. The disc here is an arbitrary space homeomorphic to the unit closed disc in the 2-dimensional Euclidean space.
I have two questions:
- The proof simply proves the homeomorphism between $\alpha$ and $\alpha'$ by claiming they are both homeomorphic to the unit closed interval $[0,1]$, why?
- If we have the homeomorphism between $\alpha$ and $\alpha'$, how can we extend it to $\alpha\cup \gamma$ and $\alpha'\cup \gamma'$?
Best Answer
$A$ is homeomorphic to disc via $f:A\to D^2$ homeomorphism. $\gamma$ is homeomorphic to $[0,1]$ by definition. And so $f(\gamma)$ is an arc in $S^1$ (the boundary of $D^2$, i.e. one dimensional sphere, a.k.a. circle). I then assume that $\alpha$ is defined as $f^{-1}(\overline{S^1\backslash f(\gamma)})$. Complement of any connected subset of $S^1$ is again connected. If it is additionally proper (like in our case) then it is homeomorphic to an interval, because proper connected subsets of $S^1$ correspond to connected subsets of $\mathbb{R}$ via stereographic projection. Finally being compact implies that it has to be a closed, bounded interval, e.g. $[0,1]$.
So the main observation here is that all arcs here have endpoints, and the intersection of $\alpha$ and $\gamma$ is precisely those two endpoints. What do I mean by endpoint? We fix $f:[0,1]\to\alpha$ homeomorphism, and then $f(0)$ and $f(1)$ are endpoints of $\alpha$. Analogously for $\gamma$. Note that this doesn't depend on the choice of $f$, although the order matters (i.e. which one is the begining, and which one is the end).
So we have a homeomorphism $F:\alpha\to\alpha'$. We also have two homeomorphisms $g:\gamma\to [0,1]$ and $g':\gamma'\to [0,1]$. Now since $[0,1]\to [0,1]$, $t\mapsto 1-t$ is a homeomorphism as well, then I can choose $g$ and $g'$ in such a way that endpoints of $\alpha$ coincide (in appropriate order) with endpoints of $\gamma$. And analogously endpoints of $\alpha'$ coincide with endpoints of $\gamma'$.
In that situation the homeomorphism $H:\alpha\cup\gamma\to\alpha'\cup\gamma'$ is given by
$$H(x)=\begin{cases} F(x) &\text{if }x\in\alpha \\ g'^{-1}\big(g(x)\big) &\text{if }x\in\gamma \end{cases}$$
The only problematic points are those that belong to $\alpha\cap\gamma$, meaning endpoints, which is why we had to choose $g$ and $g'$ correctly. The function is continuous by pasting lemma, and it has (continuous) inverse
$$H^{-1}(x)=\begin{cases} F^{-1}(x) &\text{if }x\in\alpha' \\ g^{-1}\big(g'(x)\big) &\text{if }x\in\gamma' \end{cases}$$
by the same argument.