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I think this is easy, but I'm very stuck in this problem.

Question: Let $M$ be a connected smooth manifold without boundary with $\dim M >1$, and $a_1,a_2,b_1,b_2$ $\in M$ are different points. Then there are two paths $\gamma_1,\gamma_2:[0,1]\to M$, satisfying:

  • $\gamma_1$ and $\gamma_2$ are smooth curves.
  • $\gamma_1([0,1])$ $\cap$ $\gamma_2([0,1])$ $=$ $\varnothing$.
  • $\gamma_1(0) =a_1, \gamma_1(1) = b_1$ and $\gamma_2(0) = a_2, \gamma_2(1)= b_2$.

I tried to demonstrate this problem using the following procedure.

First of all, I showed that there is a smooth function $\gamma_1: [0,1] \to M\setminus \{a_2,b_2\}$, such that $\gamma_1(0) = a_1$ and $\gamma_1(1) = b_1$, after this I tryied to show that $M\setminus\gamma_1([0,1])$ is connected, what would imply that $M\setminus \gamma_1([0,1]))$ is a connected manifold and therefore path connected , but I failed miserably.

Can anyone help me?

If possible I would like to know a bonus information (doesn't need to be proven): Does this theorem holds for an arbitrary number points?

Matheus Manzatto
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1 Answers1

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You have proved the existence of a smooth curve $\gamma_1$ from $a_1$ to $b_1$.

First, remove the self-intersections: for instance, cover $\gamma_1([0,1])$ by finitely many domains of charts (by compacity) and replace $\gamma_1$ by a piecewise linear map (one piece per domain), if it still has self-intersections then there has to be finitely many, and you can remove them by deleting the unnecessary detours made by the curve. Finally, smooth up the corners.

Now, your curve deformation retracts onto a point, which implies that $M\setminus\gamma_1([0,1])$ is just as connected as $M\setminus\{\text{one point}\}$ is.

Another way to say it is that you can consider a tubular neighbourhood $N$ of the embedded curve $\gamma_1((0,1))$, capped off by half balls around $a_1$ and $b_1$; $\partial N$ is connected (thanks, when $\dim M=2$, to the half circles around $a_1$ and $b_1$, which is where you need $\partial M=\emptyset)$, so if you consider another curve $\gamma_2$ joining $a_2$ and $b_2$, you can easily replace the part between the first time it hits $\partial N$ and the last time it leaves it by a curve that lies entirely in $\partial N$, and hence avoids $\gamma_1([0,1])$.

Edit: this question is close to another one which is quite an interesting exercise as well:

Let $f:[0,1]\to[0,1]^2$ and $g:[0,1]\to[0,1]^2$ be continuous maps such that $$\cases{f(0)=(0,0)\\f(1)=(1,1)\\g(0)=(0,1)\\g(1)=(1,0)}$$ Prove that $\operatorname{Im}f$ and $\operatorname{Im}g$ necessarily intersect.

Arnaud Mortier
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  • Sorry, but I didn't understand why "your curve deformation retracts onto a point, which implies that $M\setminusγ_1([0,1])$ is just as connected as $M\setminus{\text{one point}}$ is." can you detail more, please? – Matheus Manzatto Jun 07 '18 at 02:56
  • @MatheusManzatto If you are not used to this I added another point of view. – Arnaud Mortier Jun 07 '18 at 03:03
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    @MatheusManzatto : Do you recall that, for $x \in \mathbb{R}^2$, $\mathbb{R}^2 \smallsetminus {x}$ deformation retracts to an $S^1$ and this generalizes to $\mathbb{R}^n \smallsetminus {x}$ deformation retracts to $S^{n-1}$, which are connected for $n>1$? Notice that $\gamma_1$ is a closed, connected, compact subset of $M$, so can be deformation retracted to one of $\gamma_1$'s points relative to: that point and all of $M$ outside of a regular neighborhood of $\gamma_1$. ("Relative to ..." as in "relative homotopy".) – Eric Towers Jun 07 '18 at 03:08