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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 $\partial M$.

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  • $\begingroup$ 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. $\endgroup$ – Charlie Frohman Nov 28 '18 at 16:06
  • $\begingroup$ @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 $V'=M\backslash U'$. $\endgroup$ – RBega2 Nov 28 '18 at 21:20
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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.

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