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The mobius strip $M$ is topologically distinct from the cylinder $S^1\times I$ where $I$ is a finite segment of $\mathbb{R}$ (namely, one cannot be deformed into the other without cutting and pasting). Yet, too my understanding, they have the same homology groups:

$$H_2(M, \mathbb{Z}) \cong\{0\} \cong H_2(S^1\times I, \mathbb{Z})$$ $$H_1(M, \mathbb{Z}) \cong \mathbb{Z} \cong H_1(S^1\times I, \mathbb{Z})$$ $$H_0(M, \mathbb{Z}) \cong \mathbb{Z}\cong H_0(S^1\times I, \mathbb{Z})$$

(This calculation for the mobius strip is given here)

What tool is algebraic topology is used to distinguish orientable surfaces from non-orientable surfaces then?

Ayodan
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  • Anything homotopy invariant (e.g. homology) won't detect anything since you can deformation retract onto the core circle. What is usually used to detect orientability is the first Steifel-Whitney class, a characteristic class (certain invariants of vector bundles). Specifically, $w_1$ vanishes iff the bundle is (real) orientable, so $\mathbb R , \tilde \times , S^1$ (the twisted product, i.e. the mobius bundle) is nonorientable since it has $w_1 \neq 0$ while $\mathbb R \times S^1$ has $w_1 = 0$. – Brevan Ellefsen Sep 24 '23 at 06:38
  • Technically we need to then restrict down to $S^1 \times I$, but it's easy to do that while keeping orientability. You could alternatively exhibit the fact $\mathbb R \times S_1$ is orientable from the definition using the fact it has a nonvanishing section (a circle of nonzero height) and then use intermediate value theorem to show any section of the (finite) mobius bundle must be zero somewhere. – Brevan Ellefsen Sep 24 '23 at 06:41
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    Also see here: https://math.stackexchange.com/q/3711646/269764. (To expand on prior comments: note $S^1 \tilde \times I$ is a manifold with boundary, and such manifolds are defined to be orientable iff their interior is orientable. The interior is $S^1 \tilde \times (0,1) \cong S^1 \tilde \times \mathbb R$ so we can use characteristic classes) – Brevan Ellefsen Sep 24 '23 at 06:46
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    You can use local homology groups - a purely alg-top tool - to distinguish the two. See the post Brevan linked – FShrike Sep 24 '23 at 10:51
  • Great thank you! – Ayodan Sep 24 '23 at 14:43
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    You could also take the relative (co)homology $H_(M, \partial M)$. Another way to interpret the same invariant is as the compactly supported cohomology (or the Borel-Moore homology) of the interior $H_c^(M \setminus \partial M)$. – ronno Sep 25 '23 at 12:47

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