Is a solid sphere with a "bubble" in the middle topologically the same as a torus?

Question

Suppose there was a solid sphere or ball with radius 2 such that a sphere with radius 1 was removed from the center making a hollow cavity. A sphere with a "bubble in it" if that's easier to visualize.

Is this topologically equivalent to a torus? to a sphere? to a whole other class of shapes I'm unaware of?

Answer 1

First, Note that A sphere $\cong$ $S^n$ which is the set of points of distance equals to 1 from the origin. While a Ball $\cong$ $D^n$ which is the set of all points of distance $\le 1$ from the origin. They are different, so be careful while describing them.

I think the following argument answers your problem.

Let $X$ denotes the space you're talking about.

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This might be a clearer argument to show that $X\not\cong T_1$, where $T_1$ denotes the torus, but will require some knowledge from algebraic topology.

Choose $x_0\in X$ and a continuous map $\alpha:I\to X$ s.t. $\alpha(0)=\alpha(1)=x_0$ (loop). You can always find a way to continuously deform it to a single point, so the fundamental group $\pi(X,x_0)$ is trivial, suggesting that $X$ is simply connected

However, a loop in $T_1$ that enclose the center hole cannot be deformed to a point, suggesting that $T_1$ is not simply connected. Therefore $X\not\cong T_1$.

More information, please see here and here.

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But, $X$ can be deformation retract to a sphere $S^2$ through a continuous map $h_t:X\to A\cong S^2$ and the inverse is an inclusion map. So $X$ is homotopy equivalent to $S^2$ i.e. $X\simeq S^2$. This concept is different from homeomorphism. Definition of homotopy

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to have a better understanding about those concept mentioned above, I recommend you to read Algebraic Topology by Allen Hatcher. You could found it on website http://gen.lib.rus.ec/

Answer 2

I am not quite sure what you are really asking, so here are some arguments.

$X=\mathbb{B}^3 \setminus \frac{1}{2}\mathbb{S}^2$ has two connected components, while both the sphere and the torus have one. Hence $X$ is not homotopy equivalent (the weakest notion of topologically equivalent I am aware of) to either of them. It is however to $\mathbb{S}^2 \sqcup *$.

$Y=\mathbb{S}^2 \setminus \frac{1}{2}\mathbb{S}^2 = \mathbb{S}^2$ by definition, since $ \frac{1}{2}\mathbb{S}^2$ is not a subset of $\mathbb{S}^2$.

$Z = \mathbb{B}^3 \setminus \frac{1}{2}\mathbb{B}^3$ is homotopy equivalent to the sphere $\mathbb{S}^2$.