Möbius strip in non-orientable surface So I am trying to go over the proof of classification of surfaces and along the way, I would like to prove most result that are commonly used. So far, we can suppose the existence of a triangulation. Let establish the different definition that I would like to use here.

Definition: A compact surface $X$ without boundary is said to be orientable if and only if $H_2(X, \mathbb{Z}) \neq 0$. If $X$ is not orientable, then it is said to be unorientable.

Let $X$ be a surface and let $\mbox{Mob}$ be a Möbius strip, which is the quotient space $$\mbox{Mob} := (\mathbb{R}/2\mathbb{Z}  \times [-1, 1] )/\langle \tau \rangle.$$
where $\tau :\mathbb{R}/2\mathbb{Z}  \times [-1, 1] \to \mathbb{R}/2\mathbb{Z}  \times [-1, 1]$ is given by $t(s,t)=(s+1,-t)$. The core curve of $\mbox{Mob}$ is the simple close curve given by $\{(x,0)| x \in \mathbb{R}/2\mathbb{Z}\} $. We can also assume the existence of a PL neighborhood for our curve.

Definition: A curve $\gamma$ is one-sided if there exist an embedding $\varphi : \mbox{Mob} \to X$ such that the $\gamma$ is the core curve of $\varphi (\mbox{Mob})$.

Using these definitions, I would like to prove the following proposition:

Proposition: A surface $X$ is orientable if and only if it does not contain any one-sided curve.

Here's a proof that if $X$ is orientable, then it does not contain any one-sided curve. However, I did not find a satisfactory proof of the converse. I would ideally like to have a net proof of this result using these definitions and most importantly without using the classification of surfaces (in particular the concept of genus).
$(\Rightarrow)$ Suppose $e: \mathbb{R}/\mathbb{Z} \to X$ is an embedding, and let 
$f: \mbox{Mob} \to X$ be such that $f \circ c= e$. 
Define $M= f(\mbox{Mob})$ and let $V= X-f (\mbox{Mob}_{\frac{1}{2}})$
where 
$$ \mbox{Mob}_{\frac{1}{2}} = (\mathbb{R}/2\mathbb{Z}  \times [-\frac{1}{2}, \frac{1}{2} ] )/\langle \tau \rangle.$$
It is straightforward to construct a deformation retraction
from $M$ to $e(\mathbb{R}/\mathbb{Z})$. In particular, $H_i(M) \cong H_i(\mathbb{R}/\mathbb{Z})$.
From the inclusions $M \cap V \to X$, $\iota_M: M \cap V \to M$,
and $\iota_V: M \cap V \to V$, we obtain the Mayer-Vietoris long exact sequence
$$ \cdots \to H_2(M) \oplus H_2(V) \to H_2(X)~ \stackrel{\delta}{\to} H_1(M \cap V) \stackrel{\iota_M \oplus \iota_V}{\longrightarrow} H_1(M) \oplus H_1(V) \to \cdots.$$
The space $V$ is an open 2-dimensional manifold and hence $H_2(V) = \{0\}$. 
In addition, $H_2(M) \cong H_2(\mathbb{R}/\mathbb{Z})=\{0\}$, and so the map $\delta$ is an injection.
On the other hand, $H_1(M \cap V)$ is generated by the 1-cycle $\partial M$.
The map $\iota_M$ is induced by the inclusion into $M$, and the class of $\partial M$ 
in $H_1(M)$ is nonzero. (Indeed, in $H_1(M)$, we have $[\partial M] = 2 \cdot [c(\mathbb{R}/\mathbb{Z})]$ and  $[c(\mathbb{R}/\mathbb{Z})]$
generates $H_1(M)$.) Therefore $\iota_M$---and hence $\iota=\iota_M \oplus \iota_V$---is injective.
Since the sequence is exact, the image of $\delta$ equals the kernel of $\iota$,
and therefore $H^2(X) = \{0\}$. 
 A: We need a more local definition of orientability to answer your question. One way to do this is to say that for any point $p$ on an $n$-manifold $M$, a local orientation at $p$ is choice of a generator $g_p$ of the relative homology group $H_n(M, M \setminus p)$ (which is isomorphic to $\Bbb Z$ by excision). A global orientation on $M$ is then choice of an orientation at $x$ for every $x\in M$ so that the choice is "consistent", in the sense that for any point $p \in M$ there is a chart around $p$ containing a ball $B$ of finite radius such that all the orientations $g_x$ for $x \in B$ are images of one single generator $g_B$ of $H_n(M, M \setminus B)$ by the isomorphism  $H_n(M, M \setminus B) \to H_n(M, M \setminus x)$ induced from the inclusion $(M, M \setminus B) \hookrightarrow (M, M \setminus x)$.
There's a curious construction you could do using this machinery. Namely, consider the set $\widetilde{M}$ of all local orientations at all the points of $M$. There's a projection map $f: \widetilde{M} \to M$ that sends each local orientation to the point it orients, i.e., $f(g_p) = p$. Clearly every fiber of $f$ has cardinality two, because there are exactly two local orientations possible at each point on the manifold ($\pm 1$ are the only possible generators of $\Bbb Z$).  Give $\widetilde{M}$ the topology generated by the basis of sets of the form $\mathcal{U}(g_B)$ consisting of orientations $g_x$ which are images of the generator $g_B$ of $H_n(M, M \setminus B)$ by the map $H_n(M, M \setminus B) \to H_n(M, M \setminus x)$ for balls $B$ of finite radius on $M$. This makes $f$ into a two fold covering map. $\widetilde{M}$ is known as the "orientation double cover"
Notice that local orientations of $M$ at a point $x$ are exactly same as a fiber $f^{-1}(x)$ of this covering projection. A global orientation is a section/trivialization of the orientation double cover. There's a natural morphism $H_n(M) \to \Gamma$ where $\Gamma$ is the $\Bbb Z$-module generated by the global sections of the orientation double cover; this is given by sending each homology class $\alpha \in H_n(M)$ to the "section" $s(x) = \alpha_x$ where $\alpha_x$ is image of $\alpha$ by the homomorphism $H_n(M) \to H_n(M, M \setminus x)$. $s$ is not a section of the orientation cover because $\alpha_x$ is not necessarily a generator of $H_n(M, M \setminus x)$, but it is a multiple of a section of the orientation cover. This is in fact an isomorphism (Hatcher Chapter 3.3., Lemma 3.27). 
If $M$ is not orientable, $\widetilde{M}$ does not admit a global section ($\Gamma \cong 0$) which immediately implies $H_n(M) = 0$ (and vice versa). If it is orientable, $\widetilde{M}$ admits a global section which generates $\Gamma$. The morphism $\Gamma \to H_n(M, M \setminus x) \cong \Bbb Z$ for any $x \in M$, sending a section to it's value on the fiber $f^{-1}(x)$ is an isomorphism irrespective of the chosen $x$, as $M$ is connected. Hence, orientability of $M$ implies $H_n(M) \cong \Bbb Z$. This explains why your definition is equivalent to this one.

Suppose $M$ is a closed surface not containing embedded Moebius strips. Pick a point $p$ on $M$ and define an orientation on it by choosing a generator $g_p$ of $H_2(M,M \setminus p)$. For any other point $q$ on $M$, choose a path $\gamma$ joining $p$ and $q$, take a tubular neighborhood $U_\gamma$ of the path and consider the diagram $$H_2(M, M \setminus p) \leftarrow H_2(M, M \setminus U_\gamma) \rightarrow H_2(M, M \setminus q)$$ where both of the arrows are induced from the inclusion maps $(M, M \setminus U_\gamma) \hookrightarrow (M, M \setminus p)$ and $(M, M \setminus U_\gamma) \hookrightarrow (M,  M \setminus q)$. By a long exact sequence argument, you can argue these are both isomorphisms. So push the generator of $H_2(M, M \setminus p)$ to $H_2(M, M \setminus q)$ using the sequence of arrows in this diagram. This gives an orientation $g_q$ at $q$.
To prove that this orientation is canonical we must verify that it does not depend on the path $\gamma$ chosen from $p$ to $q$. Suppose $\gamma'$ is another path from $p$ to $q$ that gives the orientation $-g_q$ at $q$ in the above process. This means the loop $\gamma' \cdot \gamma^{-1}$ based at $q$ is "orientation reversing", i.e., transports orientation from $g_q$ to $-g_q$. This means $\gamma' \cdot \gamma^{-1}$ lifts to a path on the orientation double cover $\widetilde{M}$ joining the two preimages in the fiber $f^{-1}(p)$. Taking $V = U_\gamma \cup U_{\gamma'}$ to be a neighborhood of this loop (where $U_{\gamma'}$ is a tubular neighborhood of $\gamma'$, similarly), we can say that this means $f$ restricts to a connected orientation double cover $f : f^{-1}(V) \to V$ of $V$, i.e., $V$ is non-orientable.
Surfaces are smoothable, so giving $M$ a smooth structure, we can invoke tubular neighborhood theorem to say that $V$ is topologically an interval bundle over a circle. There are only two such objects - $S^1 \times (0, 1)$ and the (open) Moebius strip, only the latter of which admits a connected orientation double cover. This means $V$ is homeomorphic to a Moebius strip, contradicting hypothesis on $M$. Thus $g_q$ does not depend on the chosen path $\gamma$, and we could use the technique to canonically push the chosen orientation $g_p$ to an orientation $g_x$ on every point $x \in M$ by a path to have a consistent global orientation on $M$. Thus, $M$ is orientable.
