A surjective map $S^{2} \longrightarrow S^{2}$ Is there any continuous surjective map  $S^{2} \longrightarrow S^{2}$ such that it sends one of the meridians of the first sphere into the south pole of the second one?
I can see that is true, but I'm having a hard time figuring out a proper "analytical" definition (with some kind of coordinates etc.).
EDIT: For a meridian I mean the shortest arc from the north pole to the south pole (with poles included).
 A: I will prove that the following map meets the requirements:

$$\cases{X=2xy\\Y=-2yz\\Z=2y^2-1}\qquad\text{if $y>0$}$$  and
  $$\cases{X=0\\Y=0\\Z=-1}\qquad\text{otherwise}$$

In spherical coordinates (with the usual convention $0\leq \theta\leq\pi$ and $0\leq\phi\leq 2\pi$): $$f:(\theta,\phi)\mapsto \cases{(2\theta,\phi)& if $\theta\leq \frac{\pi}{2}$\\(\pi,0) & otherwise}$$
sends the equator to the south pole.
If you want to send a meridian to the SP, all you need is to send it first to the equator, and then apply the above map.
In Cartesian coordinates, the rotation represented by the following matrix sends the meridian of equation $\cases{y=0\\x^2+z^2=1}$ to the equator $\cases{z=0\\x^2+y^2=1}$:
$$M=\left(\matrix{1&0&0\\0&0&-1\\0&1&0}\right)$$
Now 


*

*Start with Cartesian $\left(\matrix{x&y&z}\right)$.

*Apply $M$ to get $\left(\matrix{x&-z&y}\right)$. 

*Turn to spherical: $$\theta =
   \arctan\left(\frac{\sqrt{x^2 + z^2}}{y}\right),
  \varphi = \arctan\left(\frac{-z}{x}\right)$$ (note that this change of variables formula is compatible with the
convention $0\leq\theta\leq \pi$ on the northern hemisphere only, but
this is sufficient for our purposes) 

*Apply the map $f$ to get $$\theta =\cases{ 2\arctan\left(\frac{\sqrt{x^2 + z^2}}{y}\right) & if $y>0$\\ 
  \pi&otherwise}$$

*Turn back to Cartesian $$\cases{X=\sin\left(2\arctan\left(\frac{\sqrt{x^2 + z^2}}{y}\right)\right)\cdot\cos\left(\arctan\left(\frac{-z}{x}\right)\right)\\Y=\sin\left(2\arctan\left(\frac{\sqrt{x^2 + z^2}}{y}\right)\right)\cdot\sin\left(\arctan\left(\frac{-z}{x}\right)\right)\\Z=\cos\left(2\arctan\left(\frac{\sqrt{x^2 + z^2}}{y}\right)\right)}\qquad\text{if $y>0$}$$

*Apply $\sin 2t=2\sin t\cos t$ to get the announced formula.

A: I thought of a map that maybe works.
Let $S^{2} \subset \mathbb{R}^{3}$ be the unit sphere in the space, with center $(0,0,0)$. I put on this the following coordinates:
$t \in [-1,1]$ indicates a point on the line from $(-1,0,0)$ to $(1,0,0)$. If I fix a $t$ of this set, I consider the circle with center $(t,0,0)$, parallel to the $yz$ plane and with radius $1-|t|$. Now, I take $\theta \in [0,2\pi]$ as the angle with the line passing trough the center and parallel to the $z$ axis in the $xz$ plane (such that the point $(0,0,-1)$ would have coordinates $(t,\theta)=(0,0)$, the point $(0,0,1)$ would have $(t,\theta)=(0,\pi)$ and so on.).
Now, the map I was looking for could be
$$f:(t,\theta)\mapsto \cases{(t,2\theta)& if $0 \leq \theta \leq \pi$\\(t,2\pi) & if $\pi \leq \theta \leq 2\pi$.}$$
