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 $\partial M$.  
 A: 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.
