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I already asked the question here but it was in a too specific way, so I asked it again here for more visibility. How do we prove the following statement :

Let $M$ be a smooth connected manifold without boundary of dim $\geq 2$ and $(x_1, y_1,... , x_n,y_n )$ be $2n$ distinct points of $M$. Then there exist smooth curves $\gamma_i :[0,1] \rightarrow M$ such that $\gamma_i (0) =x_i$ and $\gamma_i(1) =y_i$ for all $i=1,...,n$ which don't intersect each other.

Aloizio Macedo
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Sov
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2 Answers2

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Given two points $x,y$ and a finite set $F$ of points not containing $x,y$ on $M$, then there exists a smooth curve connecting $x$ and $y$ which does not intersect the finite set. This is due to the fact that $F$ is closed, hence $F^c$ is open and, therefore, a smooth connected manifold by itself (it is easy to see that a dimension $\geq 2$ manifold minus a finite set of points must be connected).

Applying the above argument recursively yields the result. You will only need the fact that a manifold of dimension $\geq 2$ minus a finite set of embedded curves is connected. By induction, it suffices to show that a manifold of dimension $\geq 2$ minus an embedded curve is connected. One way to see this is via Alexander duality*. Since $H^{n}(C)=0$ and $H^{n-1}(C)=0$ (here we use the fact that $n \geq 2$), it follows that $H_0(M,M-C)=0$ and $H_1(M,M-C)=0$, and hence that $H_0(M) \simeq H_0(M-C)$ by the long exact sequence of the pair $(M,M-C)$.

*Please, comment if you see an easier way to prove this.

To be extremely explicit, let me quote the exact duality I'm using here, since it may not be well-known (for example, we are not assuming $M$ to be orientable or compact etc).

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with the crucial observation:

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Since $C$ is an embedded submanifold, the "Cech" cohomology which Bredon is taking coincides with the singular cohomology, making the computations valid. Specifically, we are taking $K=C$, $L=\emptyset$ and $G=\mathbb{Z}/2\mathbb{Z}$ in the above theorem.

Aloizio Macedo
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  • Thanks for your answer ! The problem is that I don't know the tools you used. My professor told me that it was possible to prove it by a more classical argument (by using the smoothness of the curves). I'm looking for 3 days now and still don't find... – Sov Sep 04 '17 at 14:07
  • @Sov Hello! I'm sorry for the usage of heavy tools. I only used them because I can't see a way to justify the assertion "$M-C$ is connected" directly (actually I thought about one which uses only excision and tubular neighbourhood, thus avoiding duality, but still using homology). However, this is expected to be somewhat non-trivial, since there are results similar to this one (for example, that the sphere minus an embedded disk is connected) which are non-trivial. I'd be glad to hear a more elementary proof. – Aloizio Macedo Sep 05 '17 at 02:15
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I would recommend try proving this by induction. Proving that a connect manifold is path connected by a smooth path would be a very useful lemma.

user357980
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