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Let $D^2$ be the closed disk on the plane.

First we pick an arbitrary point $x\in bd(D^2)$ on $D^2$, and define $X = D^2-\{x\}$.

Then define another space $Y$ by removing a homeomorphic image of the closed interval $I$ from the boundary, that is, $Y = D^2-h(I)$.(The original question defines $Y$ by removing the upper closed semi-circle of $D^2$ while I think this can be generalized.)

I tried to construct the homeomorphism between two spaces but failed. The two spaces are both connected, non-compact and convex so I also failed to prove they are not homeomorphic.

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  • $\begingroup$ The boundary-punctured disc is homeomorphic to the closed upper half plane, and if you delete an arc, you instead get the upper half plane with $(-\infty, t_0) \cup (t_1, \infty)$ where $t_0 < t_1$. By scaling and translation, you see that the result of deleting any arc from the boundary of $D^2$ is homeomphic to any other. This still leaves the question of point vs arc open, but that is true (I just need to come up with the argument...) $\endgroup$
    – user98602
    Jul 18, 2018 at 2:13

2 Answers 2

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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 $X$.

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The answer of @PaulFrost is awesome, and should be the accepted answer. I thought I'd just give an explicit description of his approach.

We'll write the unit disk as $$ \mathbb{D} = \{re^{\pi it}\mid 0\le r\le1,\ -1\le t\le1\}$$

And define the lower hemisphere as $H=\{e^{\pi it}\mid -1\le t\le0\}$, and the point $q=-1$.

We have a continuous map $h:\mathbb{D}\rightarrow\mathbb{D}$, given by $$ h(re^{\pi it}) = re^{t-r+r|t|}$$ It should be checked this is well-defined (plugging in $t=1$ and $t=-1$). It's also straightforward that $h^{-1}(q)=H$. Thus we get a map $h:\mathbb{D}\setminus H\rightarrow \mathbb{D}\setminus\{q\}$.

An explicit inverse can be given too: $$ h^{-1}(\rho e^{\pi i\theta}) = \rho e^{(\rho+\theta)^2/(\rho+\theta+\rho|\rho+\theta|)}$$

One has to worry about that denominator being zero. The quadratic formula shows that happens when $\rho+\theta=0$. If $\rho<1$, then $-1<\theta<0$, and it is easy to check the fraction in the exponent goes to zero. So $h^{-1}(\rho e^{-\pi i\rho})=\rho$.

When $\rho=1$, that's a prpblem, because then we have $\theta=\pm1$. Luckily we are ignoring that point ($q$), so we get a well-defined map $$ h^{-1}:\mathbb{D}\setminus \{q\}\rightarrow\mathbb{D}\setminus H$$ and so $h$ is the desired homeomorphism.

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