I want to prove that the unit sphere $S^2$ is simply connected. In order to do this I am given the following steps:
1. Let $x_1,x_2 \in S^2$ and $\gamma \in P(S^2;x_1,x_2)$ be a path. Let $p \in S^2 \setminus \{x_1,x_2\}$. Assume $\gamma$ is not surjective. Prove that $\gamma$ is path homotopic to a path $\gamma'$ that doesn't cross $p$. (You may assume that $S^2 \setminus \{p\}$ is contractible)
2. Let $x_1,x_2 \in S^2$ en $\gamma \in P(S^2;x_1,x_2)$ again be a path. Prove that $\gamma$ is path homotopic to a non-surjective path $\gamma'$. (Has something to do with a covering space of $S^2$)
3. Prove that every loop in $S^2$ with basepoint $x_0$ is path homotopic to the constant loop with image $x_0$.
4. Conclude that $S^2$ is simply connected.
In the first step I suppose you just have to choose a point $x_3 \in S^2$, which is not on the shortest path from $x_1$ to $p$ or $p$ to $x_2$ in order to construct a path that doesn't cross $p$ and is path homotopic to $\gamma$. But the rest of the steps I don't really get. Could you please give me some hints/information to give me something to work with as I am stuck?
Thanks in advance!
Best Answer
For 2: I suspect the hint was to do with a covering of $S^2$, not a covering space ($S^2$ is simply connected, so is its own universal cover). You can choose a covering of $S^2$ by two sets: $S^2 \setminus \{x_1\}$ and $S^2 \setminus \{x_2\}$. Looking at their preimages under the path $\gamma$ gives you two open sets that cover $I$.
Now use the Lebesgue number lemma to find an open covering of $I$ by sets of the same radius, and use compactness of $I$ to find a finite subcover of this. This should let you conclude that $\gamma$ is homotopic to a "well-behaved" path - for example, a path that is a concatenation of a finite number of segments of great circles on $S^2$, whose endpoints you know.
For 3: use the earlier parts of the question.