Question about orientable manifolds Let $M$ be a connected orientable smooth manifold.
Is it true that $M$ must have only 2 orientations?
If yes, why?
 A: I think that this question can be easily answered if one analyzes the definitions carefully. Let me recall the following relevant definitions:
Definition 1 An $n$-form on a smooth $n$-manifold $M$ is a function $\Omega$ that assigns to each $p\in M$ an alternating $n$-tensor $\Omega_p$ on the tangent space $T_p(M)$. The $n$-form $\Omega$ on $M$ is said to be smooth if $\Omega(X_1,\dots,X_n)$ is a smooth function on $U$ whenever $X_1,\dots,X_n$ are smooth vector fields on any open subset $U$ of $M$.
Exercise 1: Prove that the set of alternating $n$-tensors on an $n$-dimensional vector space $V$ is itself a vector space of dimension $1$.
Exercise 2: The definition of smoothness of an $n$-form on a smooth $n$-manifold $M$ given in Definition 1 is "coordinate-free" (a desirable situation in differential geometry). Prove that this definition is equivalent to the following one using local coordinates on the smooth manifold $M$:
Definition 2 The $n$-form $\Omega$ on the smooth manifold $M$ is smooth if and only if $\Omega(E_1,\dots,E_n)$ is a smooth function on $U$ whenever $(U,\phi)$ is a coordinate neighborhood of $M$ and $E_1,\dots,E_n$ are the corresponding coordinate frames on $U$ (i.e., $E_i=\phi_{*}^{-1}\left(\frac{\partial}{\partial x^i}\right)$ for all $1\leq i\leq n$).
Definition 3 A smooth $n$-manifold $M$ is said to be orientable if there is a smooth $n$-form $\Omega$ on $M$ which is nowhere zero on $M$. In this case, $\Omega$ is said to be an orientation of $M$. An oriented smooth manifold $M$ is a smooth manifold $M$ equipped with an orientation of $M$. Two orientations $\Omega_1$ and $\Omega_2$ of $M$ are said to be equivalent, written $\Omega_1\equiv \Omega_2$, if there is a positive smooth function $f$ on $M$ such that $f\Omega_1=\Omega_2$. 
Lemma The relation $\Omega_1\equiv \Omega_2$ is an equivalence relation on the set of orientations of $M$. If the smooth manifold $M$ is connected, then there are exactly two equivalence classes.
Proof. The reflexivity of the relation follows from the smoothness of the constant function $1$ on $M$, the symmetry follows from the smoothness of $\frac{1}{f}$ on $M$ when $f$ is a positive smooth function on $M$, and the transitivity follows from the smoothness of the product $fg$ on $M$ when $f$ and $g$ are positive smooth functions on $M$.
Let $\Omega_1$ and $\Omega_2$ be orientations of $M$. Let us define a function $f$ on $M$ by the rule $f=\frac{\Omega_2}{\Omega_1}$ (this definition is valid because of Exercise 1 and the fact that $\Omega_1\neq 0$ on $M$). 
Exercise 3: Prove that $f$ is smooth on $M$. 
Note that $f\neq 0$ on $M$ because $\Omega_2\neq 0$ on $M$. Since $M$ is connected, it follows that either $f$ is positive on $M$ or negative on $M$. We conclude that either $\Omega_2\equiv \Omega_1$ or $\Omega_2\equiv -\Omega_1$. Hence there are at most two equivalence classes. However, we clearly cannot have $\Omega_1\equiv -\Omega_1$. The proof is complete.
I mention the following definition because it is very important in the formulation of Stokes' theorem and in the study of Lie groups (to mention a couple of famous examples).
Definition 4 A volume element on a smooth oriented connected manifold $M$ is a nowhere vanishing smooth $n$-form on $M$ which belongs to the equivalence class of $M$ determined by the orientation of $M$. 
I hope this helps!
A: I am also trying this question and again like you do, I am only given very basic definitions of orientations using atlas. Let me try first and see if this works, and give me your comments! (By the way I am reading sharpe's book)
Let $M$ be a connected orientable smooth manifold.
We say that an atlas $(U_{i},\phi_{i})$ is oriented if the jacobian of the transformation between two charts at the intersection is positive. (i.e. $\det\mbox{Jac}\phi_{j}\circ\phi_{i}^{-1}>0$)
Let us say that two oriented atlas $(U_{i},\phi_{i})$ and $(V_{j},\psi_{j})$ are equivalent if the union forms an oriented atlas for $M$. 
Claim
Let $(U_{i},\phi_{i})$ and $(V_{j},\psi_{j})$ be oriented atlases such that its union forms an atlas for $M$. If there exists $p\in U_{\alpha}\cap V_{\beta}\subseteq M$ such that the determinant of the Jacobian $\psi_{\beta}\circ\phi_{\alpha}^{-1}$ is positive, then both atlases are equivalent.
Proof
We will use the fact that $M$ is connected! We define two subsets as below
$S_{1}=\{p\in M: \exists U_{i},V_{j}\mbox{ such that }\det\mbox{Jac}(\psi_{j}\circ\phi_{i}^{-1})|_{p}>0\}$ 
$S_{2}=\{p\in M: \exists U_{i},V_{j}\mbox{ such that }\det\mbox{Jac}(\psi_{j}\circ\phi_{i}^{-1})|_{p}<0\}$ 
Note that we don't have $0$ as a case because $\psi_{j}\circ\phi_{i}^{-1}$ is a diffeomorphism on the intersection.
Now we have $S_{1}\cap S_{2}=\emptyset$ because suppose we have $p\in S_{1}\cap S_{2}$, then $p\in U_{i}\cap V_{j}$ and $p\in U_{k}\cap V_{l}$ such that $\det\mbox{Jac}(\psi_{j}\circ\phi_{i}^{-1})>0$ and $\det\mbox{Jac}(\psi_{l}\circ\phi_{k}^{-1})<0$. But we will run into problems because while $\det\mbox{Jac}(\psi_{j}\circ\phi_{i}^{-1})>0$, we also have
$\det\mbox{Jac}(\psi_{j}\circ\phi_{i}^{-1})=\det\mbox{Jac}(\psi_{k}\circ\psi_{l}^{-1})\det\mbox{Jac}(\psi_{l}\circ\phi_{k}^{-1})\det\mbox{Jac}(\phi_{k}\circ\phi_{i}^{-1})<0$.
On the other hand, $S_{1}\cup S_{2}=M$, because if $p$ lies in $M$, it must go into some $U_{i}$ and $V_{k}$ and so the determinant of the jacobian is either $>0$ or $<0$. 
Now it is easy to see that both $S_{1}$ and $S_{2}$ are open because taking jacobian, and then the determinant are all continuous functions and both $(0,\infty)$ and $(-\infty, 0)$ are open in $\mathbb{R}$, (so the pre-image is open in the intersection $U_{i}\cap V_{j}$ and hence is also open in $M$.
But there exists a point which both atlas agree (or compatible), this means $S_{1}$ is not empty, and so $S_{2}$ is empty because $M$ is connected.
In a same manner, if both oriented atlases disagree on one point, they disagree everywhere.
Attempted proof to your question
Since $M$ is orientable, we have already found one oriented atlas $(U_{i},\phi_{i})$.
We take the same atlas and give a negative orientation. This can be easily done by post composing $\phi_{i}$ with the operation $L:(x_{1},\ldots,x_{n})\mapsto (-x_{1},\ldots, x_{n})$ to get $L\circ\phi_{i}$.
These are the only two orientations by our previous claim because if we choose another oriented atlas $(W_{k},\lambda_{k})$ for $M$, just choose one point $p\in U_{i}\cap W_{k}$. Then because the determinant of the jacobian is either positive or negative, therefore $(W_{k},\lambda_{k})$ must agree with either $(U_{i},\phi_{i})$ or $(U_{i},L\circ\phi_{i})$.
Comments?
