You split a sphere in half at the equator and glue back together the boundaries with the function f:$S^1 \rightarrow S^1$ that maps $z$ to $z^3$. What is the fundamental group of this space?
The exact statement of the question:
Cut a sphere $S^2$ through its equator and glue it back using the attaching map f:$S^1 \rightarrow S^1$ defined by $f(z) = z^3$ . What is the fundamental group of this space?"
We take $S^1$ as the set of complex numbers with norm 1.
With Van kampen's theorem I compute the fundamental group as the trivial group. However, consider the path $\gamma$ on the equator starting at some $x_0$ and moving $2\pi/3$ degree around the equator. This path is correspond to a loop in the identification space. I fail to see how this path is homotopic to the trivial path.
Best Answer
Take the northern hemisphere $N$ and the southern hemisphere $S$. Identify $z \in \partial N$ with $z^3 \in \partial S$. Then the path $c(t) = \exp(\frac{2\pi i}{3} t), t \in [0, 1]$, in the northern hemisphere gets identified with the path $h(t) = \exp(2\pi i~t)$ in the southern hemisphere which is a loop. In fact, it's a loop that traverses the equator of the southern hemisphere. Imagine that it's a rubber band, and it's attached to $S$ at the point $z = 1$, but everywhere else it's coated in grease. Now take the part of the rubber band near $z = -1$ and press it southward. It'll slip down, pass over the south pole, and eventually contract up to the point that you've held fixed at $z = 1$. And that's your homotopy.