The "semisphere" with a horizontal segment is homeomorphic to $S^1$? Let
$$
X=(\{(x,y,z)\in \mathbb{R}^{3}: x^2+y^2+z^2=1\}\cap \{z>-1/2\})\cup\{(x,0,0):x\in[-1,1]\}
$$
be the topological space with the subspace topology of $\mathbb{R}^{3}$, graphically represented by a "semisphere" with a segment in the ecuator plane joining $(-1,0,0)$ and $(1,0,0)$.

Intuitively and attending to the graphic, I suppose I can "continuously deform" the space onto

i.e, onto $Y=(\{(x,y,z)\in \mathbb{R}^{3}: x^2+y^2+z^2=1\}\cap \{z>0\})\cup\{(x,0,0):x\in[-1,1]\}$. If a have this, it is clear that $X$ is homeomorphic to the open disc with a segment joining two points of the boundary (take the evident projection as the homeomorphism):

And the open disc with a segment joining two points of the boundary i.e the open disc with two points in the boundary is homeomorphic to $S^{1}$? I suppose some kind of homeomorphism can be constructed, but I don't know how to do it...
I am searching for some hints in two items:

*

*Is $X$ homeomorphic to $Y$? If so, how can I construct explicitly an homeomorphism between them?

*The disc with two points in the boundary is homeomorphic to $S^1$?
Thanks in advanced!
 A: Let
$$
X=(\{(x,y,z)\in \mathbb{R}^{3}: x^2+y^2+z^2=1\}\cap \{z>-1/2\})\cup\{(x,0,0):x\in[-1,1]\}
$$
There is a continuous deformation in
$$
  Y =(\{(x,y,z)\in \mathbb{R}^{3}: x^2+y^2+z^2=1\}\cap \{z\gt 0\})\cup\{(x,0,0):x\in[-1,1]\}
$$
So $X$ is homotopically equivalent to Y but they are not homemorphic since X is an open set(is the entire space,considerng it with the topology of subspace of $\Bbb R^3$ as you specified),but Y is not open in X since if you consider ${(,0,0):∈[−1,1]}$ there is no $\epsilon$ such that $B_{(1,0,0)}(\epsilon)\subset Y$since it would contain points with $z<0$,informally the ball take points under the equator for every $\epsilon$.
If I have not misunderstood even the "A=open disc with the segment" is not homeomorphic to $S^1$,since $S^1$ is compact and A not.
I agree with you that there is a deformation from X to $S^1$ and so $\pi_1(X)=\Bbb Z$ but be homotopically equivalent does not help you to say that 2 spaces are homeomorphic. In general you use such invariant to prove that 2 spaces are not homeomorphic.
