Exercise from Munkres, and a surjective loop from $[0, 1]$ onto $S^2$ I'm working through some Munkres exercises in preparation for an exam, and found this one (59.2) that I'm unsure about.
"Criticize the following 'proof' that $S^2$ is simply connected: Let $f$ be a loop in $S^2$ based at $x_0$. Choose a point $p$ of $S^2$ not lying in the image of $f$. Since $S^2 - p$ is homeomorphic with $\mathbb{R}^2$, and $\mathbb{R}^2$ is simply connected, the loop $f$ is path homotopic to the constat loop."
The only issue I can find with this argument, i.e. the only thing I'm not convinced is kosher, is the assumption that such a $p$ exists. If there existed a loop in $S^2$ whose image was all of $S^2$, this move would be illegal. Though I can imagine such a space-filling curve existing, I haven't the first clue how it'd be constructed, and am agnostic on whether such a curve could exist. Is this the problem? Or am I supposed to be noticing some other flaw? If the answer to the former is yes, how would one construct such a loop?
Thanks in advance.
 A: You have correctly identified the problem with the purported proof.
Suppose $h:[0,1]\to[0,1]\times[0,1]$ is a standard space-filling curve such as the Hilbert curve. Then you can get a space-filling loop on the 2-sphere by scaling the result of $h$ up a bit and interpreting the scaled outcome as spherical coordinates:
$$ t \mapsto \begin{pmatrix} \sin(\pi x)\cos(2\pi y) \\
\sin(\pi x)\sin(2\pi y) \\ \cos(\pi x) \end{pmatrix} \text{ where } (x,y)=h(|2t-1|)$$
The $|2t-1|$ in the argument to $h$ makes sure the curve ends back at the starting point.
A: Pick a standardly-constructed space-filling $c: I \to I\times I$ (in the sense that $c(0)$ and $c(1)$ belong to the boundary of $I \times I$). Let $\pi: I \times I \to I \times I / \sim$ be the quotient map identifying the boundary, and $h: I \times I/ \sim \to S^2$ be a homeomorphism. Then $h \circ \pi \circ c$ does the job.
A: Any Peano continuum (compact metrisable connected and locally connected) is the continuous image of $[0,1]$. And any such continuous image (if Hausdorff) is a Peano continuum. $S^2$ is one. So a curve without such a $p$ exists.
