Does there exist a continuous onto function from $S^2$ to $S^1$? Here $S^2$ is 2 dimensional unit sphere in $\mathbb{R^3}$ and $S^1$ is unit circle in $\mathbb{R^2}$. Does there exist a continuous onto function from $S^2$ to $S^1$? Till now, I am able to construct a continuous map from $S^2$ to closed unit disk in $\mathbb{R^2}$( from projection map from $\mathbb{R^3}$ to $\mathbb{R^2}$). Continuous map from closed unit disc to unit circle is defined in the following way: First we construct an onto map $f$ from closed unit disc to $\mathbb{R}\cup \{\infty\}$. $f(x,y)=\frac{1}{\sqrt{x^2+y^2}-\frac{1}{2}}$ if $\sqrt{x^2+y^2}\neq\frac{1}{2}$ and $f(x,y)=\infty$ if $\sqrt{x^2+y^2}=\frac{1}{2}$. As we know that $\mathbb{R}\cup \{\infty\}$ is homeomorphic to unit circle  we can construct an onto continuous function from closed unit disc to unit circle using $f$. The composition of function from $S^2$ onto unit disc, and function from unit disc to circle is continuous onto function. Is my reasoning correct? If not, can one construct an onto continuous function from $S^2$ to $S^1$?
 A: Sure. It must be nullhomotopic, however. For example, project the unit sphere in $\Bbb R^3$ onto the $z$-axis, thereby surjecting to the interval $[-1,1]$. Now follow with a projection from $\Bbb R$ to $S^1$ that sends the interval $[-1,1]$ at least once around the circle (say $f(t) = e^{4\pi it}$).
A: Any map $(x,y,z)\rightarrow e^{2\pi i f(x,y,z)}$ where $f:\mathbb{R^3}\rightarrow \mathbb{R}$ is a continuous function and the image $f(S^2)$ contains a closed interval of length greater than or equal to 1. For example, $f(x,y,z)=4x+4z$ or $f(x,y,z)=x^2$.
A: 
Since each map is surjective the composition is also a surjective map
A: Your reasoning is correct, but there are easier ways.
For instance, by taking the line segment from $(0,0,1)$ to $(0,0,-1)$ and project the whole sphere onto that. Then you can take that line, stretch it, and wrap it around the circle. A concrete formula might look like this:
$$
(x, y, z) \mapsto (\cos(\pi z), \sin(\pi z))
$$
Since $z$ on the unit sphere in $\Bbb R^3$ ranges from $-1$ to $1$, the argument in the trigonometric functions ranges from $-\pi$ to $\pi$, which means that the entire circle is covered.
A: Take a (continuous) map $f:\mathbb{R}^3 \to \mathbb{R}$ whose restriction to $S^2\subset \mathbb{R}^3$ is nonconstant. Then $f(S^2)$ contains an interval, and so the projection $\overline{f}:S^2 \to \mathbb{R}/t\mathbb{Z} = S^1$ is surjective for some real $t$.
