TITLE: Existence of a continuous surjective map from the surface of a sphere to a circle?
QUESTION [5 upvotes]: Does there exist a continuous surjective mapping  $f : S^2 \to S^1$ ? Intuitively, I think not, because every time in my head I try to demonstrate such a mapping, I always end up with a "tearing" in $S^2$. But I'm not sure how to prove that there doesn't exist such a mapping.

The first approach I thought of follows something like this outline:

Pick a path $\gamma$ in $S^1$ that is not homotopic to the constant path.
Artificially "lift it" to a path in $S^2$ (in some similar manner to how you life paths to covering spaces)
Show that the lifted path is homotopic to the constant path, and hence the initial path is homotopic to the constant path. Contradiction!

It doesn't seem like this way is fruitful however, since it isn't apparent how to "artificially lift the path" since $f$ isn't given as a covering map. 
I feel like this comes down to a more basic fact however, and so I attempted to make the problem easier by generalising:

If $X$ is simply connected and $f : X \to Y$ is a continuous surjective map, then $Y$ is simply connected

because that seemed to be at the core of my thinking. But I couldn't prove this either, and I couldn't find mention of a theorem like this in my textbook.

REPLY [2 votes]: Just to make the map given in @Arthur's answer explicit, consider the map $f \colon S^2 \to S^1$ given by $$f(x,y,z) = e^{2\pi i z}.$$
Here, we view $S^2 \hookrightarrow \mathbb{R}^3$ and $S^1 \hookrightarrow \mathbb{C}$ in the natural way (precisely, as the unit spheres centered at the origin in $\mathbb{R}^3$ and $\mathbb{C} \cong \mathbb{R}^2$, respectively).
This is a continuous map since it is the restriction of the continuous map $\mathbb{R}^3 \to S^1$ given by the same formula. It is not difficult to see that this map is also surjective.
(It is in fact also a smooth surjection from $S^2$ to $S^1$, if we view $S^2$ and $S^1$ as smooth manifolds.)