Prove there is no single parametrization of a circle I'm trying to prove that there is no single parametrization of the circle. In other words, there is no bijective smooth map with a smooth inverse between the circle and an open interval.
Intuitively, I think of deforming an open interval into a circle. Because the interval is open, there is always a gap between the two ends of the interval and if I make them overlap, it's not a bijection. Alternatively, I think of the interval $(a,b)$ and that if f is smooth and maps onto a circle, $f(a+\varepsilon)$ and $f(b-\varepsilon)$ must be close to each other but $a+\varepsilon$ and $b-\varepsilon$ are distant.
I can't figure out how to write that formally though and would appreciate help in drafting that proof.
 A: Your intuition is correct. In fact smoothness is not required: there is no homeomorphism (continuous bijection with continuous inverse) between an open interval and a circle (with the topology induced by the circle's embedding in the plane). The reason is that the circle is compact and the open interval is not, and homeomorphisms preserve the property of a space of being compact.
A: You said you don't know what a homeomorphism is, so here is an explanation at a lower level. In particular it doesn't use the language of topology, just calculus.
Let us abbreviate "bijective smooth map" as "nice map". Note that the composition of two nice maps is a nice map.
Now, the standard parameterization $(\cos t, \sin t)$ shows that there is a nice map from an open interval to the unit circle with $(-1, 0)$ deleted. In fact the inverse of this map is also a nice map. So by composition any nice map onto the unit circle with $(-1, 0)$ deleted will give a nice map onto an open interval.
This is the key.
Suppose for contradiction we have a nice map $f$ from some interval $(a, b)$ onto the unit circle. Now remove the point $(-1, 0)$. There must be a number $\theta$ in $(a, b)$ such that 
$f(\theta) = (-1, 0)$. What remains is a nice map from $(a, \theta) \cup (\theta, b)$ to the unit circle with that point deleted.
And by our earlier reasoning, this gives a nice map from $(a, \theta) \cup (\theta, b)$
onto an open interval, say $(c, d)$. 
But a nice map from $(a, \theta) \cup (\theta, b)$ onto $(c, d)$ is impossible. For the graph of such a nice map will be a smooth curve that goes from one "half" of a rectangle to the other without crossing the line between them, and this is geometrically impossible. Making this rigorous would require more sophisticated language, but by drawing a picture you'll see what I mean.
Smoothness is not used at any point in this argument; we don't use smoothness of the curve in the rectangle, we only use that it's unbroken AKA continuous. So if we change "nice map" to simply mean "bijective continuous map" then everything still holds.
