If spacetime is $\mathbb{R}^4$, the space of all possible connections of the $G$-bundle is connected. But consider what happens when we remove a timelike worldline from spacetime. The resulting spacetime is homotopically equivalent to the topological sphere $S^2$.
Is this above rigorous or just handwaving?
Does this say that $\mathbb{R}^4$ removing a line $\mathbb{R}^1$, we get the compliment topology as a 2-sphere $S^2$?
This does not make much sense to me. Can you explain this fact?
My attempt: Suppose we cut out the tubular neighborhood of $\mathbb{R}^1$, then we consider
$$
\mathbb{R}^4 \smallsetminus D^3 \times \mathbb{R}^1
$$
The remaining topology is still a 4-manifold. If we consider beginning with a sphere topology and cut out a tubular neighborhood of $S^1$, for the compact and finite size topology, I get:
$$
S^4 \smallsetminus D^3 \times S^1 = S^2 \times D^2
$$
Then we may interpret the later as the $S^2$ topology but we still need to product it with $D^2$.
- Does my attempt explain all that Wikipedia paragraph about to say?
"Homotopy equivalence" is a weaker property than "topological equivalence" (homeomorphism). This concept is explained in many places, for instance Hatcher's Algebraic Topology, so I won't explain it here.
In your decomposition, $S^2\times D^2$ deformation retracts onto $S^2\times\{(0,0)\}$, and deformation retraction is one way we obtain homotopy equivalences.
In fact, $S^4\setminus S^1$ deformation retracts onto $S^4\setminus D^3\times S^1$, so these spaces are homotopy equivalent as well.
Transitivity of homotopy equivalence implies then that $S^4\setminus S^1$ is homotopy equivalent to $S^2$.