Suppose one removes a line and a point from $\mathbb{R}^3$ (the line not intersecting the point). What is the second homotopy group of the resulting space?
My thoughts: the fact that a point is missing gives rise to the usual set of maps that wrap the two-sphere around the point an integer number of times. So there must be a $\mathbb{Z}$ subgroup. However, one can also stretch the sphere around the missing line before wrapping it over the point, and this seems to give rise to an inequivalent mapping from the two-sphere to the space.
Indeed, one can stretch the sphere around the line any integer number of times before wrapping it around the point, and since $\pi_2$ is always Abelian, my guess is that
$$\pi_2 = \mathbb{Z} \times \mathbb{Z} \,.$$
However, I don't know how to check whether this is correct, nor do I know of any general techniques to attack this sort of problem. Any help in these regards is much appreciated.
It's a good start but the answer is a bit more complicated than that.
Show that your space deformation retracts on the wedge sum of a circle (that goes once around your line) and a $2$-sphere (that goes once around your point).
Show that the universal cover of that wedge sum is an infinite chain of ...sphere-line segment-sphere-line segment... where the fibre is therefore $\Bbb Z$.
From the homotopy exact sequence you see that $\pi_2(X)$ is the free abelian group on infinitely (countably) many generators.