Proving directly that $S^2$ is simply connected: is a surjective loop homotopic to a non-surjective one? I want to prove that the sphere $S^2$ is simply connected. The following argument appeared at Munkre's book on topology and it is "flawed", since there are surjective loops in the sphere $S^2$. But I heard that it is possible to show that such a surjective loop is homotopic to a non-surjective loop and hence the argument can be used again to finally prove that $S^2$ is simply connected. 
The argument is the following:
Let $f:I = [0,1] \rightarrow S^2$ be a loop in $S^2$. Assuming that $f$ is not surjective, let $p\not\in f(I)$. Now considering $\varphi: S^2 \setminus\{p\} \rightarrow \mathbb R^2$, the stereographic projection at $p$, we obtain a loop in $\mathbb R^2: \varphi \circ f.$ Since $\mathbb R^2$ is simply connected, we obtain that $\varphi\circ f$ is path homotopic to a constant path given by the path homotopy $H$. Hence, $\varphi^{-1}\circ H$  is a path homotopy in $S^2$ between $f$ and a constant path.
Now the question is: is it true that a surjective loop in $S^2$ is path homotopic to a non-surjective loop in $S^2$? How to prove that? Thank you!
 A: It's just a Lebesgue number argument. I thought this argument would be in Munkres "Topology", he is usually good on Lebesgue number arguments, but I don't have my copy to check.
Anyway, here is a version of the argument. 
Let $\{U_i\}$ be the open cover of $S^2$ by open hemispheres. Using the given loop $f : [0,1] \to S^2$ we obtain an open cover $\{f^{-1}(U_i)\}$ of $[0,1]$. Let $\lambda>0$ be a Lebesgue number for the latter open cover. Choose an integer $n \ge 1$ such that $\frac{1}{n} < \lambda$. Subdivide the interval into subintervals of length $\frac{1}{n}$,
$$[0,1] = [x_0,x_1] \cup [x_1,x_2] \cup \ldots \cup [x_{n-1},x_n]
$$
It follows that for each $i=1,...,n$ the path $f| [x_{i-1},x_i]$ is contained in an open hemisphere of $S^2$. Let $\delta_i$ be the unique great circle path in that open hemisphere having the same endpoints as $f | [x_{i-1},x_i]$. Since open hemispheres are homomorphic to $\mathbb R^2$, it follows that $f | [x_{i-1},x_i]$ is path homotopic to $\delta_i$. Thus $f$ is path homotopic to the concatenation $\delta_1 * \ldots * \delta_n$. 
Now convince yourself that a union of finitely many great circle subpaths of $S^2$ is not surjective in $S^2$.
