Determine whether the fundamental group of $\mathbb{R}^3$ with its non-negative half-axes removed is trivial, infinite cyclic, or isomorphic to the figure eight space. I found this answer:
Why do we have that $\alpha*\beta=\gamma$? I can't see how we have this homotopy or deformation.
PS: I think we are actually supposed to solve this by showing that we can find that the figure eight space is a deformation retract of this space, or homotopy equivalent. Do you see a way of doing this? I cannot really see how to define the deformations.
Here's one approach to the question in the postscript:
If we denote by $X$ the subspace of $\Bbb R^3 - \{ 0 \}$ whose fundamental group we are computing, one can show that the mapping $\Bbb R^3 - \{ 0 \} \to \Bbb S^2 \subset \Bbb R^3$ defined by $x \mapsto \frac{x}{||x||}$ restricts to a homotopy between $X$ and a sphere with three points deleted, namely $Y := \Bbb S^2 - \{(1, 0, 0), (0, 1, 0), (0, 0, 1)\}$, and this restriction and the inclusion $Y \hookrightarrow X$ together comprise a homotopy equivalence $X \simeq Y$. The thrice-punctured sphere $Y$ is then homeomorphic (via, e.g., stereographic projection from one of the deleted points) to the plane with two points deleted, $\Bbb R^2 - \{p, q\}$, and this space is in turn homotopic to the figure eight space (see page 3 of these notes of Munkres for some diagrams that indicate how to write down explicitly this latter homotopy equivalence).