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:
Is $\Bbb R^2\times\Bbb S^2$ 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!
Best Answer
If a space $X$ is simply connected, then each space $Y$ which is homeomorphic to $X$ is also simply connected. A stronger result is that if $Y$ is homotopy equivalent to $X$, then $Y$ is simply connected. See Proving that the fundamental groups of two spaces with same homotopy type are isomorphic .
Now $\mathbb R^2 \times S^2$ is homotopy equivalent to $S^2$. But it is well-known that $S^2$ is simply connected.
You can also invoke the fact that $\pi_1(A \times B) \approx \pi_1(A) \times \pi_1(B)$. But $\pi_1(\mathbb R^2) = 0$ and $\pi_1(\mathbb S^2) = 0$.