Complement of codim 2 manifold is connected? If M is a connected n-manifold and N is a codim 2 submanifold
then M-N is connected.
Is this true?
I want to show $H_0(M-N)=\mathbb Z$. I think it's better to do in $\mathbb Z_2$ coefficient because we get orientability. But because no compactness is mentioned I can't use Poincare duality. How can I resolve this?
 A: This has little to do with orientability, nor compactness, Poincare duality, nor $\mathbb Z_2$ coefficients. It's just proving path connectivity of $M-N$. Also, the proof works in any codimension $q \ge 2$, and I'll write it that way.
Consider any $x,y \in M-N$.
Start by choosing a continuous path $\gamma : [0,1] \to M$ such that $\gamma(0)=x$ and $\gamma(1)=y$. Of course the path $\gamma$ might hit $N$.
The intuition is that because $N$ is of codimension $q \ge 2$, when the path $\gamma$ comes close to $N$ you can push it away. The result is a path $\delta : [0,1] \to M-N$. The endpoints $\gamma(0)$ and $\gamma(1)$ will be far enough away that they don't get moved when you push other portions of the path away from $N$, hence $\delta(0)=\gamma(0)$ and $\delta(1)=\gamma(1)$, and we obtain a path $\delta$ in $M-N$ from $x$ to $y$
To make this intuition work one applies the Tubular Neighborhood Theorem and the same Lebesgue number lemma techniques that are used elsewhere in algebraic topology.
Here are some details. The Tubular Neighborhood Theorem applied to the submanifold $N \subset M$ produces an open $U \subset M$ with $N \subset U$, and a continuous retract $f : U \to N$, such that $f$ is a fiber bundle in which the fibers are homeomorphic to the open q-dimensional disc $D^q$. Each $p \in N$ has a path connected open neighborhood $V_p \subset N$ and a homeomorphism $h_p : U_p = f^{-1}(V_p) \approx V_p \times D^q$ such that $h_p(U_p \cap N) = V_p \times \{0\}$. Because both $V_p$ and $D^q - \{0\}$ are path connected (this is where we use $q \ge 2$), it follows that $h_p(U_p - N) = V_p \times (D^q - \{0\})$ is path connected. Because $h_p$ is a homeomorphism, it follows that $U_p - N$ is path connected.
You can choose $U$ to be a subset of any prechosen open neighborhood of $N$; let's choose $U$ to be a subset of $M - \{x,y\}$.
Consider the following open covering of $M$:
$$\mathcal U = \{M-N\} \cup \{U_p \mid p \in N\}
$$
By pullback we get an open covering of $[0,1]$:
$$\gamma^{-1}(\mathcal U) = \{\gamma^{-1}(M-N)\} \cup \{\gamma^{-1}(U_p) \mid p \in N\}
$$
Applying the Lebesgue number lemma, we obtain a subdivision
$$0 = x_0 < x_1 < ... < x_K = 1
$$
such that for each $k=1,\ldots,K$ the restriction $\gamma[x_{k-1},x_k]$ has image contained in some element of the open covering $\gamma^{-1}(\mathcal U)$.
Using all this structure, let's construct a path $\delta : [0,1] \to M-N$ with $\delta(0)=x$, $\delta(1)=y$.
Step 1: If $\gamma(x_k) \not\in N$, let $\delta(x_k)=\gamma(x_k)$. And if $\gamma[x_{k-1},x_k]$ is disjoint from $N$, let $\delta(x)=\gamma(x)$ for all $x \in [x_{k-1},x_k]$. Note that this case applies whenever $\gamma[x_{k-1},x_k]$ contains a point that is not in $U$, because $M-N$ is the only element of the open cover $\mathcal U$ that has nonempty intersection with $M-U$. So, for example, $\delta$ equals $\gamma$ on the subintervals $[x_0,x_1]$ and $[x_{K-1},x_K]$.
Step 2: For any $x_k$, if $\gamma(x_k) \in N$ then choose $\delta(x_k)$ to be a point of $U_{\gamma(x_k)} - N$ such that $f(\delta(x_k)) = \gamma(x_k)$.
Step 3: For any $k=1,...,K$, if $\gamma[x_{k-1},x_k] \cap N \ne \emptyset$ then there exists $p \in N$ such that $\gamma[x_{k-1},x_k] \subset U_p$. In Steps 1 and 2 we have already found endpoint values $\delta(x_{k-1})$, $\delta(x_k)$ in $U_p - N$. Since $U_p - N$ is path connected, we can define $\delta$ on $[x_{k-1},x_k]$ with those endpoint values.
A: Here is an argument which works for a connected $n$-dimensional topological manifold $M$ and a closed subset $C\subset M$  homeomorphic to a manifold of dimension $\le n-2$. I am not sure what happens if you drop the assumption that $C$ is closed. My suspicion is that the complement need not be path-connected, but it might still be connected.
First of all, it suffices to work with oriented manifolds $M$ (otherwise, you work in the orientation cover of $M$). Next, using only the assumption that $C$ is a closed subset of $M$, a form of the Poincare-Lefschetz duality in this situation yields an isomorphism
$$ 
\check{H}^*_c(C)\cong H_{n-*}(M, M-C),
$$
where the left hand side is the Chech cohomology with compact support. For manifolds, Chech cohomology is isomorphic to the singular cohomology and, hence, the dimension assumption yields
$$
\check{H}^{n-1}_c(C)={H}^{n-1}_c(C)=0.
$$
Next, I will use the long exact sequence of the pair $(M, M-C)$:
$$
\to H_1(M)\to 0=H_1(M, M-C)\to \tilde{H}_0(M-C) \to \tilde{H}_0(M)=0
$$
Thus, $\tilde{H}_0(M-C)=0$ and, hence, $M-C$ is path-connected.
A: The goal of this answer is to use Moishe Kohan's idea to prove the general statement for not-necessarily-closed submanifolds. In fact even more than that can be said.
Let $M$ be a metrizable topological manifold and let $f: S \to M$ be a continuous injection from a second-countable topological manifold with $\dim S \leq \dim M - 2$. (Notice that this need not be an embedding. For instance, $f$ could be a map from $\Bbb Z^2$ with image $\Bbb Q^2 \subset M$.)

Theorem: $M \setminus f(S)$ is path-connected. In fact, for any two points $x,y \in M \setminus f(S)$, and any continuous path $\gamma: [0,1] \to M$ with $\gamma(0) = x$ and $\gamma(1) = y$, there is an arbitrarily close continuous path with the same endpoints whose image is contained in $M \setminus S$.

Remark. Injectivity is necessary because of the existence of space-filling curves. Second-countability is necessary as otherwise one could take $S$ to be $M$ equipped with the discrete topology, and $f$ to be the canonical injection.


Lemma: If $D$ is a closed disc embedded in $M$ of dimension $\dim D \leq \dim M - 2$ and $x, y \in M \setminus D$, write $\mathcal P_{x \to y} M$ for the space of continuous paths from $x$ to $y$ in $M$. Then $\mathcal P_{x \to y} (M \setminus D)$ is open and dense in $\mathcal P_{x \to y} M$.

Proof: because $S$ is closed in $M$, the set of paths with image in $M \setminus D$ is open in the compact-open topology (by definition).
Denseness is more difficult. Suppose $\gamma: [0,1] \to M$ is a continuous path with $\gamma(0) = x$ and $\gamma(1) = y$. Then $\gamma^{-1}(D)$ is a closed subset of $[0,1]$ not including the endpoints.
Fix $\epsilon > 0$. We will construct a path $\eta$ with $d_\infty(\gamma, \eta) \leq \epsilon$ and $\eta^{-1}(D) = \varnothing$. Carrying out this construction for all $\eta$ proves the desired density result.
Using the Lebesgue number lemma, one may find $n$ large so that
$\gamma([i/n, (i+1)/n]$ has diameter less than $\epsilon/2$ for all $0 \leq i < n$.
Write $U_{\epsilon,i} \subset M$ for the open set of points $$\{x \in M \mid \text{sup}_{t \in [i/n, (i+1)/n]} d(\gamma(t), x) < \epsilon\}.$$
For each $0 < i < n$ pick $x_i \in U_{\epsilon,i} \setminus D$. This exists because otherwise $D$ would contain an open subset of $M$, contradicting its dimension and the invariance of domain theorem. Set $x_0 = x$ and $x_n = y$.
Using Moishe Kohan's result, for each $0 \leq i < n$ there exists a path $\eta_i: [i/n, (i+1)/n] \to U_{\epsilon, i} \setminus D$ with $\eta_i(i/n) = x_i$ and $\eta_i\left(\frac{i+1}{n}\right) = x_{i+1}$.
Notice that $d_\infty(\eta_i, \gamma|_{[i/n, (i+1)/n]}) < \epsilon.$
Now define $\eta(t)$ to be $\eta_i(t)$ whenever $i/n \leq t \leq (i+1)/n$. Notice that this gives a well-defined continuous path on all of $[0,1]$, with $$d_\infty(\eta, \gamma) < \epsilon,$$ and with $\eta$ missing $D$, as desired.

To prove the Theorem, write $S$ as the union of countably many closed discs $D_i$. Because the domain is compact and the codomain is Hausdorff, $f$ gives a homeomorphism of $D_i$ onto $f(D_i)$. We have already seen that $\mathcal P_{x \to y} [M \setminus f(D_i)]$ is dense in $\mathcal P_{x \to y} M$. But $$\mathcal P_{x \to y} [M \setminus f(S)] = \bigcap_i \mathcal P_{x \to y} [M \setminus f(D_i)].$$
This is an intersection of countably many open dense sets inside the complete metric space $\mathcal P_{x \to y} M$. Complete metric spaces are Baire, and hence this intersection is once again dense and in particular nonempty. The theorem follows.
