Simply connectedness in $R^3$ with a spherical hole?

I understand why $R^3 – {(0,0,0)}$ is simply connected, and I also understand why $R^2 – {(0,0)}$ is not simply connected. The way I look it at is if checking if the region is $a)$ path-connected and $b)$ any curve can be contracted to a point in the region.

From what I reasoned it seems there is a pattern, a hole in $R^2$ prevents simply connectedness, a missing line in $R^3$ does the same, and so I reasoned for $R^n$ any $n-2$ dimensional missing figure (or higher) would prevent the region from being simply connected.

Then I was posed with the scenario: take $D$ to be all of $3D$ space except for a sphere of radius 1, is $D$ simply connected? The answer is apparently yes, D is simply connected because “the spherical hole does not prevent paths from contracting to points while remaining in $D$”. However, now I’m confused because a spherical hole is a $3D$ hole and it goes against my previous conjecture. Also, according to this MIT video: , $R^3 -$ a circle is not simply connected.

So why is $R^3$ – a spherical hole simply connected?

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Think about loops with some fixed point in $\mathbb{R}^3$. You need to be able to contract $\it{any}$ possible loop attached to that fixed point back to the fixed point. For the spherical hole (deleted ball) you can always pass the loop around the ball. Now picture a finite length pole with a loop around it. You can still pass the loop around the rod, but not through it. If the rod were infinite in length (a line in $\mathbb{R}^3$) you could neither pull the loop through the rod or around the end of the rod.

$\mathbb{R}^3$ minus a 3-dimensional ball is simply connected.

$\mathbb{R}^3$ minus a sphere (e.g. minus the surface of a ball) is not, though for a slightly different reason than that $\mathbb{R}^3$ minus the $z$-Axis is not.

In $\mathbb{R}^3$ minus a sphere, your region consists of two disconnected parts – the inside of the sphere and the outside. You can continuously transform every path into a point, but you can not transform continuously transform every path into every other path – you can only do that if the paths reside on same side of the sphere.

Note that $A:=\mathbb{R}^{3}\setminus S^{2}$ is not connected and thus not path connected, which breaks the definition of being simply connected already. So the answer isn’t very deep.

In fact, to define $\pi_{1}(A)$, which is base-point free notation, you need $A$ to be path connected. So the notation $\pi_{1}(A)$ is already a little ambiguous. For non path-connected spaces you have to study each $\pi_{1}(A,x_{0})$ separately with $x_{0}$ being a base point at a given path component. Each of the path components of $A$ are simply connected, though.

Firstly, R^3 – {origin} is simply connected. The reason is that, naively, you have more degree of freedom as far as finding a continuous path in the region is concerned.