In my class, orientation on a $C^\infty$ manifold is defined as follows:
Let $\mathcal{O}_0$ be the canonical orientation on $\mathbb{R}^m$ (i.e. the equivalence class (with respect to orientation) that contains the canonical basis).
An orientation on a manifold $M$ with $\dim M =m$ (with or without boundary) is a family $\mathcal{O}=\{\mathcal{O}_x\}_{x \in M}$ of orientations on the tangent spaces $T_xM$ such that, for every $x \in M$, there exists a local parametrization $g:U \rightarrow g(U) \subseteq M$ (where $U$ is an open set of $\mathbb{R}^m$ or $H^m$ that contains $x$) such that $\text{d}g_u(\mathcal{O}_0)=\mathcal{O}_x$ (meaning that $\text{d}g_u([\mathcal{B}]):=[\text{d}g_u(\mathcal{B})]=\mathcal{O}_x$, where $\mathcal{B}$ is the canonical basis of $\mathbb{R}^m)$ for every $u \in U$. In this case, $g$ is called compatible.
After that, we have the following observation:
If $\mathcal{O}$ is an orientation on $M$, then also $-\mathcal{O}:=\{-\mathcal{O}_x\}_{x \in M}$ is an orientation on $M$:
Given a family of local compatible parametrizations with respect to $\mathcal{O}$, we can assume that the domains $U$ of the parametrizations are open balls centered in the origin (otherwise we can restrict the parametrization to an open ball contained in $U$ that contains $x$, and traslate the center of the ball in the origin), so composing these parametrizations with the reflection $(x_1, \dots,x_m) \mapsto (x_1,\dots,-x_m)$ we obtain a family of parametrizations that send $\mathcal{O}_0$ in $-\mathcal{O}_x$.
Until now everything is fine, but after this first observation, we have this second observation:
If $\dim M=1$ and $\partial M \neq \emptyset$, our definition doesn't work well, since, for example, the reasoning in the previous observation doesn't work in this particular case, so we modify it in the following way:
If $M$ is connected and compact and $ \partial M \neq \emptyset$, then an orientation on $M$ is a family $\mathcal{O}=\{\text{d} \varphi _t (\mathcal{O}_0)\}_{t \in [0,1]}$ where $\varphi:[0,1] \rightarrow M$ is a diffeomorphism.
I really don't understand this observation: why the previous one fails if $\dim M=1$ and $M$ has boundary? Why replace the first definition with this new one?
I tried to search for this issue in some textbooks like Tu, Milnor, Guillemin and Pollack, but none of them even mention it, and also to me the reasoning in the first observation seems to work also in this case. So where is the problem?
Best Answer
Up to diffeomorphism there are only two $1$-manifolds with boundary. These are $M = [0,1)$ and $M = [0,1]$.
Since both manifolds $M$ are subsets of $\mathbb R$, each orientation $\{ \mathcal O_x \}$ of $M$ has the property $\mathcal O_x \in \{ \mathcal O_0, -\mathcal O_0\}$ at each $x \in M$.
$M = [0,1)$ does not make any trouble, we can use the general definition of an orientation.
However, in contrast to higher dimensional manifolds it only has a single orientation given by $\mathcal O_x = \mathcal O_0$ for all $x \in M$.
To see this, take any local parametrization $g : [0,a) \to M$ with $g(0) = 0$. This map $g$ must be increasing, thus its usual derivative $g'(u)$ is everywhere positive. Since the linear map $dg_u$ is multiplication by $g'(u)$, we see that $dg_u(\mathcal O_0) = \mathcal O_0$ for all $u \in [0,a)$. In other words, $\mathcal O_x = \mathcal O_0$ in a neighborhood $[0,b) = g([0,a))$ of $0$ in $M$. We can now easily show that $\mathcal O_x = \mathcal O_0$ for all $x \in M$. Assume that
$\mathcal O_x = -\mathcal O_0$ for some $x \in [b,1)$. Let $\xi = \inf \{x \in [b,1) \mid \mathcal O_x = -\mathcal O_0 \}$. Clearly $\xi \ge b > 0$. Using a local parametrization at $\xi$, we see that in some neighborhood $V$ of $\xi$ we either have $\mathcal O_x = \mathcal O_0$ for all $x \in V$ or $\mathcal O_x = -\mathcal O_0$ for all $x \in V$. But this contradicts the definition of $\xi$: For $x \in V$ with $x < \xi$ we have $\mathcal O_x = \mathcal O_0$, and there exist $x \in V$ with $x > \xi$ and $\mathcal O_x = -\mathcal O_0$.
The manifold $M' = (0,1]$ is diffeomorphic to $M$, thus it also only has a single orientation. This is given by $\mathcal O_x = -\mathcal O_0$ for all $x \in M$. To see this, take any local parametrization $g : [0,a) \to M$ with $g(0) = 1$. This map $g$ must be decreasing, thus its usual derivative $g'(u)$ is everywhere negative.
This shows that $M = [0,1]$ cannot have an orientation in the sense of the general definition. In fact, any orientation of $[0,1]$ induces orientations on the subsets $[0,1)$ and $(0,1]$ and these are not compatible on $(0,1)$.
The reason for this phenomenon is this:
The reflection $x_1 \mapsto -x_1$ does not map $H^1 = [0,\infty)$ into itself.
As a consequence we get that $[0,1)$ and $(0,1]$ only have a single orientation which causes the above problem.
In dimensions $> 1$ the reflections are "well-behaved" so that for each orientation $\mathcal O$ also $-\mathcal O$ is an orientation.
An alternative to the adhoc-definition of an orienation for $[0,1]$ given in your class is this.
In local parametrizations allow all halfspaces of $\mathbb R^m$, that is all sets $$H^m_{i,\pm} = \{(x_1,\ldots,x_m) \mid \pm x_i \ge 0 \} .$$
For $m = 1$ we get $H^1_{1,+} = [0,\infty)$ and $H^1_{1,-} = (\infty,0]$. Then $[0,1)$ and $(0,1]$ have two orientations so that $[0,1]$ does not make any trouble.