$H_1(X,X-N,\mathbb Z_2)=\mathbb Z_2$, proof? Given $X$ a connected manifold, and $N$ a connected codimension-1 submanifold. Can someone help me show how the result of the title holds?
I tried excision. That is, I tried removing $U^c-N$ where $U$ is a trivial neighbourhood about a point in $N$, but ended up with the pair $(U\cup N, U-N)$. So I can't get anywhere. The idea is to reduce the question to computing it for $(\mathbb R^n, \mathbb R^n\setminus\mathbb R^{n-1})$ with mod 2 coefficients. For this pair it's clearly $\mathbb Z_2$ by the long exact sequence for the pair.
So I want to know how to reduce it or know a better method. Thanks!
 A: This might not be the easiest way to go around it and I'd be very happy to see a more elementary solution, but I'm sharing this since it is natural from the point of view of Thom spaces (and also proves a stronger result). Note that you cannot really hope for a "three-line proof" in the topological case, since your statement implies the separation theorem for circles in the plane.
I assume some regularity assumptions that provide the existence of a tubular neighbourhood$N \subseteq U \simeq N _{X/N}$ , for example it's enough to assume that $X, N$ are smooth and $N \hookrightarrow X$ is an embedding. By excision, we have 
$H_{i}(X, X \setminus N) \simeq H_{i}(U, U \setminus N) \simeq H_{i}(N _{X/N}, N_{X/N} \setminus N) \simeq H_{i}(DN _{X/N}, DN _{X/N} \setminus N)$,
where $DN _{X/N}$ is the associated unit disk bundle. If $SN _{X/N}$ is the associated unit sphere bundle, then $(DN _{X/N}, SN _{X/N}) \rightarrow (DN _{X/N}, DN _{X/N} \setminus N)$ is a homotopy equivalence of pairs by radial retraction. Now, the inclusion $SN _{X/N} \hookrightarrow DN _{X/N}$ is a cofibration and so we have
$H_{i}(DN _{X/N}, SN_{X/N}) \simeq \tilde{H}_{i}(DN_{X/X} / SN_{X/N})$, 
where $TN _{X/N} = DN_{X/N} / SN_{X/N}$ is by definition the Thom space of the normal bundle. Since you're working over $\mathbb{Z}_{2}$, every bundle is orientable and the Thom isomorphism theorem applies to tell us that 
$H^{i}(N) \simeq \tilde{H}^{i+1}(TN _{X/N})$,
which immediately implies your statement, since $\mathbb{Z}_{2}$ is a field and so homology and cohomology are dual. 
