Least number of embedded hypersurfaces which separate a compact manifold Suppose that $N^{n+1}$ is a connected closed smooth manifold. Define $$S(N^{n+1}) = \min \bigl\{ k \mid \forall (M^n_i), \, \bigcup_{i=1}^k M^n_i \text{ separates } N^{n+1} \bigr\}.$$ 
Here $M^{n}_i$ ($1 \leq i \leq k$) are arbitrary $k$-connected closed smoothly embedded hypersurface of $N^{n+1}$ which satisfy $M^n_i\cap M^n_j=\emptyset$ if $i\neq j$.
I want to know the relationship between $S(N^{n+1})$ and the homology group of $N^{n+1}$ or Betti number of $N^{n+1}$, especially can we give $S(N^{n+1})$ an upper bound in terms of $b_1(N^{n+1})$?
For $n=1$ and orientable case, I guess $S(\Sigma^2)=g(\Sigma^2)+1=\frac{b_1(\Sigma^2)}{2}+1$. But I don't know how to prove it strictly. And how about the higher dimension case? For simplicity, we can firstly consider orientable case.
 A: Let $N$ be a smooth closed (connected) manifold and suppose there is a collection of $k$ closed hypersurfaces $X_i\subseteq N$ which do not disconnect $N$ when removed from it.  In other words, assume $S(N)\geq k+1$.  Let $U$ be a small open neighbor of the union of the $X_i$ and let $V$ denote $N \setminus \bigcup_i X_i$.  Then $U$ and $V$ cover $N$, so we may use the unreduced Mayer-Vietoris sequence.  Further, note that $V$ is connected.
We get an exact sequence $$... \rightarrow H_1(N)\rightarrow H_0(U\cap V)\rightarrow H_0(U)\oplus H_0(V)\rightarrow H_0(N)\rightarrow 0.$$
Now, consider the part of $U$ around $X_i$ and intersect it with $V$.  I have a picture in my head (but no proof), that this deformation retracts to the orientation some double covering of $X_i$. 
In particular, it has $1$ or two components, depending on whether $X_i$ is non-orientable  has trivial normal bundle or not (thanks Mike!).  Let $l$ denote the number of orientable  $X_i$ with trivial normal bundle.
The previous paragraph gives that $H_0(U\cap V)$ is isomorphic to $k + l$.  Further, $H_0(U) \oplus H_0(V)$ is $k+1$ copies of $\mathbb{Z}$, because $V$ is connected.  Since $N$ is connected, $H_0(N)\cong\mathbb{Z}$.
So, the exact sequence really looks like $$...\rightarrow H_1(N)\rightarrow \mathbb{Z}^{k+l}\rightarrow \mathbb{Z}^{k+1}\rightarrow \mathbb{Z}\rightarrow 0.$$
Now, $\mathbb{Z}^{k+1}\rightarrow \mathbb{Z}$ is surjective, so the kernel is isomorphic to a lattice of the form $\mathbb{Z}^k$.  This is the image of $\mathbb{Z}^{k+l}$, so the map from $\mathbb{Z}^{k+l}\rightarrow \mathbb{Z}^{k+1}$ has kernel isomorphic to a lattice of the form $\mathbb{Z}^k$.
This is the image of $H_1(N)$.  Thus, we have shown that $H_1(N)$ has a quotient of the form $\mathbb{Z}^l$, and since $\mathbb{Z}$ is free, it follows that $H_1(N)$ contains a subgroup isomorphic to $\mathbb{Z}^l$.
Thus, we have shown:

Suppose $S(N)\geq k+1$, then $b_1(N)\geq l$ where is the largest number of $X_i$ which can be chosen to be orientable have trivial normal bundle.

In the case where $N$ is a oriented surface, each $X_i$ is a circle, so will automatically have a trivial normal bundle, so $l = k$ in this case, and there is a surjection $H_1(N)\rightarrow \mathbb{Z}^k$.
I claim the following:   The kernel of the map $H_1(N)\rightarrow H_0(U\cap V)$ consists of all those curves in $H_1(N)$ whose intersection number with every $X_i$ is trivial.
First, assume $\alpha\in H_1(N)$ is in the kernel.  Write $\alpha  = \beta + \gamma$ where $\gamma $ lies entirely in $U$.  Then the map $H_1(N)\rightarrow H_0(U\cap V)$ maps $\alpha$ to $\partial \gamma$, so $\partial \gamma = 0$.  It follows that $\gamma$ is homologically equivalent to a disjoint union of circles each of which is contained in a small neighborhood of some $X_i$.  But these neighborhoods are diffeomorphic to $X_i\times (0,1)$, so $\gamma$ has $0$ intersection number with each $X_i$.
Conversely, if $\alpha\in H_1(N)$ has trivial intersection number with each $X_i$, then, up to homology equivalence,  the image of $\alpha$ is disjoint from $X_i$, and thus, wlog, disjoint from $U$.  Thus, if we decompose $\alpha =\beta + \gamma$ with $\gamma \in U$, we actually have $\gamma = 0$.  Hence, $\partial \gamma = 0$, so $\alpha$ is in the kernel.
Having established the kernel of $H_1(N)\rightarrow H_0(U\cap V)$ consists of precisely those $\alpha \in H_1(N)$ which have $0$ intersection number with every $X_i$, we now note that each $X_i$ is in the kernel.  That is, since $X_i\cap X_j = \emptyset$, $X_i\cdot X_j = 0$.  Similarly, $X_i\cdot X_i = 0$ since the normal bundle of $X_i$ is trivial.
Thus, we have shown:

If $N$ is a closed orientable surface with $S(N)\geq k+1$, then $b_1(N)\geq 2k$.

Said another way, if you can find $k$ curves which do NOT disconnect your (orientable) surface, you must have $b_1 \geq 2k$.  So, $S(\Sigma_g) \leq g+1$.  But it's easy to see that $S(\Sigma_g) > g$ by drawing the relevant curves.  So we have shown $S(\Sigma_g) = g+1$.
