How can the Möbius band be an image of a parametrization if it is not orientable? I'm using a book of analysis on $\mathbb{R}^n$ with the following definitions:

*

*m-dimensional parametrization of class $C^k$ of $V\subseteq \mathbb{R}^n$: a homeomorphism $\phi:V_0\rightarrow V$ of class $C^k$, $k\geq 1$, $V_0\subseteq \mathbb{R}^n$, such that the derivative map $\phi '(x):\mathbb{R}^m\rightarrow\mathbb{R}^n$ is injective for each $x\in V_0$

*Positive overlap: two parametrizations (same dimension, same class), $\phi: V_0\rightarrow V$, $\psi:W_0\rightarrow W$ overlap positively if either $V\cap W=\emptyset$ or $\det J(\psi^{-1} \circ \phi)(x)>0$, where $J$ is the Jacobian matrix defined for $\psi^{-1} \circ \phi |_{\phi^{-1}(V\cap W)}$

*Atlas of a manifold: an atlas for a m-dimensional manifold $M\subseteq\mathbb{R}^n$ of class $C^k$ is a set $A$ of m-dimensional, $C^k$ parametrizations whose images cover $M$. An oriented atlas is an atlas $A$ such that each parametrization overlap positively with each other

*A manifold $M$ is said to be orientable if it allows an oriented atlas

With that the book reaches the conclusion that if a manifold $M$ is the image of one parametrization $\phi:V_0\rightarrow M$, then it is orientable because it allows the atlas $A=\{\phi\}$ which is oriented.
However, when the book introduces the Möbius band, for the usual purpose of an example of not orientable manifold, it starts by saying that it can be seen as the image of a parametrization $f:(0,1)\times [0,2\pi]\rightarrow\mathbb{R}^3$ given by $f(s,t)=(\cos(t)+(s-\frac{1}{2})\cos(t/2)\cos(t),\sin(t)+(s-\frac{1}{2})\cos(t/2)\sin(t), (s-\frac{1}{2})\sin(t/2))$. My question is how that can be if the Möbius band is not orientable?
What I think is happening is that the book is using the same word "parametrization" for  both the rigorous definition it gave (and used to define orientation), and the more common use of the word, but I am not sure. I can see, for example that the function $f$ described above is not bijective for the given domain, but I think if we adjust the domain to $(0,1)\times[0,2\pi)$ it becomes bijective.
If anyone would help me to see how it is possible that the Möbius strip is both not orientable but it is the image of that function, I would be very thankful.
 A: Here are two clues:

*

*The subset $(0,1) \times [0,2\pi]$ is not open in $\mathbb R^2$.

*The function $f : (0,1) \times [0,2\pi] \to \mathbb R^3$ (defined by the given formula) is not injective.

So that function $f$ is definitely not a parameterization in the sense explained in your four-bullet-point definition of a manifold. So yes, the terminology "parameterization", when used for that definition of the Möbius band, is being used in a loose manner.
So that point of that definition is really just to define the Möbius band as a subset of $\mathbb R^3$, and not (yet) to produce an atlas for it.
However, the interesting thing is that if you restrict the function $f$ to the subset $(0,1) \times (0,2\pi)$ then that restricted function is indeed a parameterization for a subset of the Möbius band as in your four-bullet-point definition. Furthermore, you can then use the same formula except on a different domain $(0,1) \times (-\pi,+\pi)$, to get another parameterization for a different subset of the Möbius band. And then, lo and behold, these two parameterizations taken together define an atlas for the Möbius band... although not an oriented atlas, which is a good thing because the Möbius band is not orientable.
