# 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 $H^2(X) = \{0\}$.

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.