Disjoint curves in a smooth manifold.

Question

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.

Answer 1

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.

Answer 2

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.

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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).

with the crucial observation:

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.