Say we have a unit circle $S^{1}$, with an equivalence relation $\sim$ defined as $(x, y) \sim (x', y') \iff y = y' =0$.
So this gives that just two opposite points of the circle are equivalent. Does this mean that the quotient space $S^{1}/\sim$ is a figure 8, since the circle is pinched into the center by joining the two points $(-1, 0)$ and $(1, 0)$?
In this case I believe that the fundamental group of $S^{1}/\sim$ is given by $\pi_{1}(S^{1}/\sim) \cong$ the free group of two generators, via the Van Kampen theorem.
If $n=2$, and we have a sphere, and indeed in higher dimensions too, is it that the equator contracts into the center, giving two spheres attached at a point?
I am fairly new to the idea of topology, so if anyone has any online sources that gives what spaces are homeomorphic, and what they strong deformation retract to, that would be helpful also.
Best Answer
Yes, the quotient is topologically a figure-8 for the case $n = 1$.
Your computation of the fundamental group for $n = 1$ is correct.
For $n > 1$, the space is indeed topologically two spheres sharing a single point.
So it seems to me that you're doing fine so far! Keep at it!