Question about connected manifold

I need some tip to prove the following:

If $$N^{n}$$ is a connected manifold and $$M^{m}$$ is a closed submanifold of $$N$$, such that $$n-m\geq 2$$, then $$N-M$$ is connected.

I am supposed to use transversality to prove this task, but I couldn't come up with any idea. If someone can give me a clue, even solving by different geometric methods, thanks.

• Hint: Use generic transversality between paths in N and the submanifold M. – Moishe Kohan May 25 at 22:01
• Yes, that was the tip I received, but I can't see how to use it – ArkPDEnational May 25 at 23:58
• OK, here is another hint: Can a curve and a submanifold of codimension $\ge 2$ have nonempty transversal intersection? – Moishe Kohan May 26 at 0:04

This answer will just flesh out Moishe Kohan's comments.

If $$N$$ is a connected manifold, $$M$$ a closed submanifold of codimension at most 2, then $$N\setminus M$$ is connected.

Proof.

Let $$p,q\in N\setminus M$$ be points. Since $$N$$ is connected, choose a smooth path $$f:[0,1]\to N$$ from $$p$$ to $$q$$ in $$N$$.

$$f$$ is transversal to $$N$$ at $$0$$ and $$1$$, (since $$f(0)=p$$, $$f(1)=q$$ and $$p,q\not\in M$$) so there is a homotopy of $$f$$ to a path $$f':[0,1]\to N$$ transversal to $$M$$ everywhere such that $$f$$ and $$f'$$ agree on some neighborhood of $$0$$ and $$1$$. Thus $$f'(0)=f(0)=p$$, $$f'(1)=f(1)=q$$, so $$f'$$ is a path from $$p$$ to $$q$$ transversal to $$M$$. Then if $$f'(t)\in M$$, for some $$t$$, then $$\dim (df'_t\Bbb{R} + T_{f'(t)}M) \le 1+m \le n-2+1 = n-1 < n,$$ so if $$f'(t)\in M$$ for any $$t$$, $$f'$$ is not transversal to $$M$$. Thus $$f'$$ is a path in $$N\setminus M$$. Hence $$N\setminus M$$ is connected. $$\blacksquare$$

A reference for these techniques: Differential Topology by Guillemin and Pollack

For given two points in the compliment, we will first find a path having a nonempty intersection with a submanifold $$M$$. And we will find a path having a empty intersection in $$\epsilon$$-disance wrt to the previous :

i) For $$p,\ q$$ in a compliment of a submanifold $$M$$ in $$(N,d)$$ where $$d$$ is a Riemannian distance, then we have a foot $$p_f,\ q_f$$ s.t. $$(l_p:=)\ d(p,M)=d(p,p_f),\ p_f\in M$$ And similar for $$q_f$$.

When $$c_p$$ is a shortest path of unit speed from $$p$$ to its foot $$p_f$$, then set $$p_f'=c_p(l_p-\epsilon )$$ for some $$\epsilon >0$$. That is, $$p_f'$$ is close to a submanifold $$M$$.

ii) When $$\alpha$$ is a shortest path of unit speed from $$\alpha(0)=p_f$$ to $$\alpha(l)=q_f$$ wrt an intrinsic metric of $$M$$, then there is a curve $$\alpha'$$ in the boundary of $$\epsilon$$-tubular neighborhood of the curve $$\alpha$$ in $$N$$ s.t. $$\alpha'$$ starts at $$p_f'$$, is in a compliment of $$M$$, and ends in the closed $$\epsilon$$-ball $$B_\epsilon(q_f)$$.

Hence $$B_\epsilon(q_f)$$ contains the end points $$p_f''$$ of $$\alpha'$$ and $$q_f'$$

iii) Since $$B_{\epsilon}(q_f)$$ is bi-Lipschitz to Euclidean ball, then there is a path from $$p_f''$$ to $$q_f'$$ in a compliment of $$M$$ in the ball $$B_{\epsilon}(q_f)$$, since $$M$$ has at least codimension $$2$$.

• Do you mean $d(p,M)$? Also I don't understand how $p_f^\prime(t)$ is defined. From what I understood $p_f^\prime$ depends on $p$ and $c_p$, I don't see how you make it a function of $t$. Finally it is not clear to me why you can find a path from $p_f$ to $q_f$ in $M$. – Adam Chalumeau May 25 at 20:27
• Yes, I couldn't understand the proof as well. If it'd be great if you can explain some details. – ArkPDEnational May 25 at 20:45