# Homology and Mobius Band

Let $$M$$ be the closed Mobius band (so $$M$$ is a two-manifold with boundary). Suppose $$U$$ is an open subset of $$M$$ so that $$\bar{U}\cap \partial M$$ is empty and the inclusion $$i:U\to M$$ induces the zero map on homology with $$\mathbb{Z}_2$$ coefficients (i.e., $$U$$ is not the tubular neighborhood of the central circle of $$M$$). Notice I'm not assuming that $$U$$ is connected or simply connected.

How do I rigorously show the following (which I feel is intuitively clear): If $$V$$ is $$M\backslash U$$, then the inclusion map $$j: V\to M$$ induces a surjective map $$j_*:H_1(V; \mathbb{Z}_2) \to H_1(M; \mathbb{Z}_2).$$ Intuitively, this means that $$V$$ contains a one-cycle homologous to the central circle which seems very plausible. I tried using Mayer-Vietoris, but my algebraic topology skills are too rusty and couldn't convince myself that it would work. Notice you have to use $$\bar{U}$$ is disjoint from $$\partial M$$ or else $$M$$ minus a transverse interval gives a counterexample (though I just realized think this might be resolved by looking at relative homology groups...).

Edit: I originally had $$V$$ be the component of $$M\backslash U$$ that contained $$\partial M$$, but realized this was false (consider a small tubular neighborhood of a parallel curve to $$\partial M$$.

• I think it might be two general to have a clean answer. The sets U and V need to be nice enough that you can use the Mayer-Vietoris sequence. – Charlie Frohman Nov 28 '18 at 16:06
• @CharlieFrohman I can see that is possible. If it matters, I'm happy to assume there is a open set $U'\subset U$ with $\bar{U}'\subset U$ and prove the same result for $V'=M\backslash U'$. – RBega2 Nov 28 '18 at 21:20

Let $$U$$ and $$V$$ be two subsets whose interiors cover $$M$$, with $$\partial M\subset V$$ and $$i_*:H_1(U;\mathbb{Z}_2)\to H_1(M;\mathbb{Z}_2)$$ being the zero map. We can deduce from the universal coefficient theorem that the image of $$i_*:H_1(U)\to H_1(M)$$ lies in $$2H_1(M)$$.
The relative Mayer-Vietoris sequence contains the exact sequence $$H_2(M,\partial M)\to H_1(U\cap V)\to H_1(U)\oplus H_1(V,\partial M)\to H_1(M,\partial M)\to H_0(U\cap V).$$ Since $$(M,\partial M)$$ is a good pair, $$H_2(M,\partial M)\cong H_2(M/\partial M)\cong H_2(\mathbb{R}\mathrm{P}^2)=0$$, and similarly $$H_1(M,\partial M)=\mathbb{Z}/2\mathbb{Z}$$. Since $$H_0(U\cap V)$$ is a free abelian group, the map from $$H_1(M,\partial M)$$ has trivial image, hence we get a short exact sequence $$0\to H_1(U\cap V)\to H_1(U)\oplus H_1(V,\partial M)\to \mathbb{Z}/2\mathbb{Z}\to 0.$$ Now, consider the quotient $$q:M\to M/\partial M$$. We have $$q_*:H_1(M)\to H_1(M,\partial M)$$ being the quotient map $$\mathbb{Z}\to\mathbb{Z}/2\mathbb{Z}$$, where a generator of $$H_1(M,\partial M)$$ is the core circle from $$H_1(M)$$ (an equally good generator: a transverse interval, in your terminology). The composition $$q_*\circ i_*:H_1(U)\to H_1(M,\partial M)$$ is the zero map, hence, by thinking about exactly what the maps in the Mayer-Vietoris sequence are, the $$\mathbb{Z}/\mathbb{2}\mathbb{Z}$$ comes entirely from the $$H_1(V,\partial M)$$. That is, $$H_1(V,\partial M)\to H_1(M,\partial M)$$ is surjective.
Now, consider the naturality of the long exact sequences for the inclusion $$(V,\partial M)\to (M,\partial M)$$.$$\require{AMScd}$$ $$\begin{CD} H_1(\partial M) @>>> H_1(V) @>>> H_1(V,\partial M) @>>> H_0(\partial M) \\ @VVV @VVV @VVV @VVV \\ H_1(\partial M) @>>> H_1(M) @>>> H_1(M,\partial M) @>>> H_0(\partial M) \end{CD}$$ The first and fourth vertical induced maps are isomorphisms, and the third we established is surjective. By one of the four lemmas, this implies $$H_1(V)\to H_1(M)$$ is surjective. From the UCT, we can deduce the weaker statement that $$H_1(V;\mathbb{Z}/2\mathbb{Z})\to H_1(M;\mathbb{Z}/2\mathbb{Z})$$ is surjective as well.