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.

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


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 $2$.

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    $\begingroup$ 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 $M$. $\endgroup$ – Adam Chalumeau May 25 at 20:27
  • $\begingroup$ Yes, I couldn't understand the proof as well. If it'd be great if you can explain some details. $\endgroup$ – ArkPDEnational May 25 at 20:45

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