Making a smooth curve that runs around the origin exactly once? I have just started to read do Carmo's book and this is the first exercise in the book:
Find a parametrised curve $\alpha (t)$ whose trace is the circle $x^2 + y^2 = 1$ such that $\alpha (t)$ runs clockwise around the circle with $\alpha (0) = (0,1)$.
My answer to it is this:
Let $\alpha (t) = (\sin t, \cos t)$ and $t \in (-1, 4 \pi)$. Then $\alpha (0) = (0,1)$ and $\alpha ({\pi \over 2} )= (1, 0)$ so it's clear that $\alpha$ runs around $(0,0)$ clockwise.
After going over this again I realised that this is not a proof that it goes around the origin clockwise. The book does not give a definition of "going around the origin clockwise". I did some digging and came across the winding number . Unfortunately, this is defined for closed curves. But the starting point and end point of my curve do not even exist as it is defined on an open interval. 
Hence my first follow up question on this exercise is:

(1) How to mathematically rigorously prove that  $\alpha (t) = (\sin
 t, \cos t)$ where $t \in (-1, 4 \pi)$ goes around the origin
   clockwise?

Now assuming that my answer is okay as it is I was wondering if it's possible to make a smooth curve that goes around the origin exactly once, that is, $\alpha : (a,b) \to \mathbb R^2$ such that $\alpha (a) = (1,0) = \alpha (b)$ and $\alpha (x) \neq \alpha (y)$ if $x \neq y$.
What troubles me is that the domain is an open interval:  The book defines smooth curves as maps defined on $(a,b) \subseteq \mathbb R$.

(2) Is it possible to define a smooth curve (image the unit circle) defined on an open interval that
  goes around the origin exactly once?

After going over this exercise several times I noticed something else that is bothering me, too:

(3) The exercise statement does not require the curve to be
  differentiable. Is this on purpose or a typo? Does it make sense to
  ask for a possibly discontinuous curve?

And lastly: 

(4) I was wondering whether if it is impossible to define a smooth
   curve $\alpha : (a,b) \to S^1$ that is bijective if it is possible to
   define a smooth bijection  $\alpha : [a,b] \to S^1$ or $\alpha : [a,b)
 \to S^1$?

 A: In the interest of answering your questions, as posed:


*

*A reasonable definition for "going around the circle clockwise" which matches intuition is that if you identify $\Bbb R^2$ with $\{(x,y,0):x,y\in\Bbb R\}\subset \Bbb R^3$, then $\mathbf x(t)\times \mathbf x'(t)=(0,0,z(t))$ for some non-positive function $z$, i.e. $z(t)\leq 0$ for all $t$. From here, I suspect you can answer the question yourself.
This isn't quite right, in that rather than "going around the circle clockwise", it's more like "going around the circle clockwise everywhere". You could probably coarsen this restriction by demanding that instead of holding for all $t$, it holds for a "sufficiently large" subset of $t$, but there's no notion of "sufficiently large" that matches my intuition.

*Honestly, the definition you give here kind of depends on which answer you believe. If you want the answer to be 'no', you can say that it means $f$ must be a bijection. If you want the answer to be 'yes', you can say that $f:(a,b)\to \Bbb R^2$ extends to a smooth map $\tilde f:(\tilde a,\tilde b)\to\Bbb R^2$ with $\tilde a<a<b<\tilde b$ such that $\tilde f\big|_{[a,b]}$ is a closed curve with winding number 1. (You may also want $f$ itself to be surjective, if you're really sweating the last point.)
The latter construction may seem more contrived, but it is actually a standard construction in differential topology.

*Actually it does: do Carmo defines a parametrized curve to be differentiable.

*The half-open interval is of course possible, your curve can be modified for an example. The closed interval is not possible, since any continuous bijection from a compact space to a Hausdorff space is a homeomorphism (it is easy to show, for instance, that it is closed). The fact that $S^1$ and $[a,b]$ are not homeomorphic follows by your favorite method.
The open interval is again impossible, but (as the attempt so far shows) it is more subtle. One approach is to take a continuous extension of $f$ to $[a,b]$: take $\tilde f(a)=\lim_{x\to a}f(x)$ and similarly for $f(b)$; these limits exist since $S^1$ is compact. But now look at some point which is not $\tilde f(b)$ but is $\varepsilon$-close to $\tilde f(a)$. This has exactly one $\tilde f$-preimage because $f$ is a bijection. Since this preimage must be $\delta$-close to $f^{-1}(\tilde f(a))\neq a$, it cannot be $\delta$-close to $a$. 
To do this rigorously you will probably have to invoke uniform continuity of $\tilde f$, but that's okay since the domain is compact. You would also have to show that you can force $\delta$ to be small enough: this should follow by sending $\varepsilon\to 0$ and noting that this would force $f$ to send an entire interval to $f(a)$.
A: Do I understand it correctly that you want an injective ($\alpha(x) \neq \alpha(y)$ if $x \neq y$), smooth curve with image the unit circle defined on an open interval? 
There does not exist such a map. There is a simple topological argument for this: if you remove a point from the interval $(a,b)$, the interval becomes disconnected, while removing a point from the circle keeps it connected. Since the function you are considering is bijective, continuous and is a function from a subset of $\mathbb{R}$ to a subset of $\mathbb{R}$, you cannot send a space that is made up of two connected components to a connected space, because there must be a bijection between connected components.
In other language, once you go around the circle once, except for hitting the point $(1,0)$, you have two options. Either you hit the point for a certain $t \in (a,b)$, meaning that where $(t,b)$ is sent, would overlap with a previous part of the curve, or you would never hit the point $(1,0)$.
Edit: The fact that we are considering a map from a subset $U$ in $\mathbb{R}$ to a subset V of $\mathbb{R}$ in the topological argument, is essential. In fact, being injective and continuous is already enough in this case, because of Invariance of Domain (https://en.wikipedia.org/wiki/Invariance_of_domain). This states that such a function is an open map and because connected components are both open and closed, such a function defines a bijection on your connected components.
