Question of whether two given spaces are homeomorphic.

Let $e : \mathbb{R} \to S^1, e(t) = e^{it}$. Then a homeomorphic image of a closed interval in $S^1$ is any set having the form $J = e([a,b])$ with $a < b$ and $b - a < 2\pi$.

We have $X = D^2 \backslash \{e(c) \}$ for some $c \in \mathbb{R}$ and $Y = D^2 \backslash e([a,b])$ for suitable $a,b$ as above. Define $r : [a,a + 2 \pi] \to [c,c+2\pi]$, $r(x) = c$ for $t \in [a,b]$, $r(x) = c + \frac{2\pi (x - b)}{a+2\pi-b}$ for $t \in [b,a+2\pi]$. This is a continuous map. Define

$$H : [a,a + 2 \pi] \times [0,1] \to [c,c+2\pi], H(x,t) = (1-t)(x -a + c) + t r(x) .$$

This is a continuous map such that (with $H_t(x) = H(x,t)$)

(1) $H_1 = r$

(2) $H_t(a) = c, H_t(a+2\pi) = c + 2\pi$ for all $t$

(3) $H_t$ is strictly increasing, in particular bijective, for all $t < 1$.

Define

$$G : [a,a + 2 \pi] \times [0,1] \to D^2, G(x,t) = te(H(x,t))$$

which is again continuous. Consider the continuous map

$$p : [a,a + 2 \pi] \times [0,1] \to D^2, p(x,t) = te(x) .$$

Since domain and range are compact, it is an identification map. It is easy to check that if $p(\xi) = p(\xi')$, then $G(\xi) = G(\xi')$. Therefore we obtain a unique continuous $g : D^2 \to D^2$ such that $g \circ p = G$. By construction $g$ maps $Y$ bijectively onto $X$. The restriction $g' : Y \to X$ is an open map which shows that it is a homeomorphism: Let $U \subset Y$ be open. Then $D^2 \backslash U$ is compact, thus $g(D^2 \backslash U) ) = D^2 \backslash g(U)$ is compact so that $g'(U) = g(U)$ is open in $X$.