Is the 3-sphere a product of the 2-sphere with the 1-sphere? I know that the n-torus is the product of n 1-spheres. I know that $S^2$ can be gotten from $T^2$ by cutting it in half and identifying the edges as one point. So is there any connection I can make between $T^3$, $S^3$, and $S^2 \times S^1$? $T^3$ requires three angles ranging $[0,2\pi)$. $S^3$ requires two angles ranging from $[0,2\pi)$ and one angle ranging from $[0,\pi]$. The exact same thing is true for $S^2 \times S^1$, so does this mean they are the same?
 A: No. We cannot learn very much from a single chart on a manifold. The coordinates you have described are not a true, well defined coordinate chart because of ambiguity when the last coordiante equals $0$ or $\pi$. But if we restrict the last coordinate to $(0,\pi)$, we do have a chart, but it doesn't describe global properties of the manifold very well.
Algebraic topology is the primary toolbox used to tell manifolds apart. The difference between these is that every closed loop in $S^3$ contracts to a point, while a closed loop in $S^2\times S^1$ of the form $\gamma(t)=(x,t)$ for $x\in S^2$ a constant and $0\leq t\leq 2\pi$ does not contract. Therefore, these two spaces must be different.
A: So Alex has a nice answer, but I thought it'd also be appropriate to mention the Hopf fibration, and fiber bundles in general.
A fiber bundle is a space that looks locally like the Cartesian product of a base space $B$ and a fiber $F$, but is different globally. One of the simplest examples of this is the Mobius strip. If you take the open interval $I = (0,1)$ and the circle $S^1$, their Cartesian product is a cylinder, $S^1 \times I$. A Mobius strip is made of the same two spaces, but there is a twist when going around the circle - a $Z_2$ invariant that distinguishes the trivial product from the fiber bundle. It's the same idea for $S^3$ and $S^2 \times S^1$ - they look the same locally, but globally they're distinguished by a quantity called the winding number.
A: Allowing Algebraic Topology in your life, things get really easy.
Look at the Homology groups of $S^3$ and $S^2 \times S^1$ . $S^2 \times S^1$ has $\Bbb Z$ as it's 0th, 1st, 2nd and 3rd Homology group and all others to be 0 whereas $S^3$ has $\Bbb Z$ as only it's 0th and 3rd Homology group and all other Homology groups to be 0 . 
I guess you are asking whether the two spaces are homeomorphic, but since they have different Homology groups they cannot be homotopic and in particular, homeomorphic!   
