The following answer states that the $\Bbb R^2\times\Bbb S^2$ space is simply connected:
However, the following post confirms that $\Bbb R^2\times\Bbb S^2$ is homeomorphic to $\Bbb R^4$ with a line removed:
Simply connected means that any path between any two points can be continuously deformed into any other path between the same points without leaving the space:
In $\Bbb R^4$, I can connect any two points by two paths, one on one side of the removed line and the other on the other side of the removed line. I do not see how one path can be smoothly transitioned into another without crossing the removed line. This is evidently impossible in $\Bbb R^3$ with a line removed, does the presence of another dimension in $\Bbb R^4$ makes it possible?
Equivalently, a space is simply connected if any loop can be contracted to a point without leaving the space. Consider a loop around the removed line. If I contract this loop to a point, this point would be on the removed line and thus outside the space. Again, self evident in $\Bbb R^3$, does the presence of an extra dimension in $\Bbb R^4$ allows contracting such a loop to a point outside the removed line without crossing it?
So is $\Bbb R^4$ with a line removed simply connected? And is the homeomorphic $\Bbb R^2\times\Bbb S^2$ simply connected as well? What am I missing? Thank you!