Is map $f:\mathbb{D}^2\to \mathbb{S}^2$ that fix $\partial \mathbb{D}^2$ surjective on a hemisphere? I have a map $f:\mathbb{D}^2\to \mathbb{S}^2$ that fix $\partial \mathbb{D}^2$. It seems intuitve believe that $f(\mathbb{D}^2)=\mathbb{S}^2_+$ or $f(\mathbb{D}^2)=\mathbb{S}^2_-$ (hemispheres north and south), but I could not to prove it.
Could you give me a clue?
Thank you
 A: Suppose there are two points not in the image of $f$, one on the North emisphere and one on the South emisphere. Removing the two points, up to homeomorphism you get a map
$$\hat{f} : D^2 \to S^1 \times \mathbb{R}$$
That maps the boundary to a generator of the fundamental group on the right. This is impossible, because an element of the $\pi_1$ extends to a disk precisely when it is trivial.
On the other side, it is possible to have a surjective map with the constraints you prescribed. Firstly, let me notice there is a surjective function $g: D^2 \to S^2$ that maps the boundary to the South Pole: this is kind of "put a disk on a sphere and then collapse the boundary to the South Pole. A similar one can be done with North Pole, that we call $g_1$.
Now define $f: D^2 \to S^2 $ as:
$$ f(x) = g(2x) ,\quad \|x \| \le 1/2$$
$$ f(x) = g_1 \left ( \frac{x}{\| x \| } \left (\|x\| - \frac{1}{2} \right ) \right ) , \quad 1/2 \le  \|x\| \le 1 $$
You can verify that it satisfy your constraints. Use yous imagination!
A: Think of $\Bbb{S}^2$ as the surface of a tennis ball and think of $\Bbb{D}^2$ as a sock of about the same diameter as the tennis ball. Put the tennis ball in the bottom of the sock with the south pole at the toes and tie a piece of string around the sock so that it neatly traps the whole tennis ball with the knotted string at the north pole. Now fold the rest of the sock into tucks so that the edge fits snugly around the equator of the tennis ball (if you choose a short enough sock you will only need one fold). This gives you a picture of a surjection of $\Bbb{D}^2$ onto $\Bbb{S}^2$ that identifies $\partial\Bbb{D}^2$ with $\Bbb{S}^1$ in the standard way.
