Every connected topological manifold has a homeomorphism taking $p_i$ to $q_i$ for any $i=1,\dots, k$.

Question

This is problem 5-4 from Lee's Introduction to Topological Manifolds.

If $M$ is a connected topological manifold and $(p_1, \dots, p_k)$ and $(q_1, \dots, q_k)$ are two ordered $k$-tuples of distinct points in $M$, then there is a homeomorphism $F: M \to M$ such that $F(p_i)=q_i$ for $i=1,\dots ,k$.

I know how to prove this in the case of $k=1$. By induction, it suffices to prove this in the case $k=2$. I think I have to use the gluing lemma to extend two homeomorphisms each taking $p_1$ to $q_1$ and $p_2$ to $q_2$, but I am lost here. I would greatly appreciate any help.

Answer 1

The point $p_1$ has an open neighborhood homeomorphic to a ball in $\Bbb{R}^k$. There is a homeomorphism dragging all of the $p_i$ to points in this neighborhood.

Consider this homeomorphism acting on $p_2$: there is an arc connecting a point in the neighbrhood of $p_1$ and $p_2$. This arc has a tubular neighbourhood, which is homeomorphic to a $k$-ball. This homeomorphism contracts that $k$-ball to a small $k$-ball entirely inside the neighborhood of $p_1$. If we are concerned about "interference" when trying to move all the $p_i$ at once, compose these $k$-ball contractions one at a time.

Likewise, there is a homeomorphism dragging all of the $q_i$ to a neighborhood of $q_1$. Finally, there is a homeomorphism taking the neighborhood of $p_1$ to the neighborhood of $q_1$. Assemble these in the order first, third, inverse of the second.

Answer 2

We will assume $\dim M>1$. (See below for more information.)

Strengthen your case $k=1$. Are you able to show that given $p,q$ and a finite amount of points $A$ such that $p,q \notin A$, then there exists a homeomorphism $F$ for which $F(p)=q$ and the points of $A$ are fixed? $^{(1)}$

If so, then for every $p_i,q_i$ do this with $A_i=\{p_1,\cdots,\widehat{p_i},\cdots p_n,q_1, \cdots, \widehat{q_i}, \cdots, q_n\}$. Then $F_n \circ F_{n-1} \circ \cdots \circ F_1$ will be your required homeomorphism.

$^{(1)}$ In order to show this, one can proceed as follows: define the following equivalence relation on $M \backslash A$: $x \sim y$ if there exists an homeomorphism $F:M \to M $ that takes $x$ to $y$ and leaves $A$ fixed. Solving the issue locally with translations by a suitable vector multiplied with a bump function, you can prove that each class is open. Since $M \backslash A$ is connected if $\dim M >1$, there can be only one class, and the result is proved.

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If I'm not mistaken, the result seems to be false if we don't impose $\dim M > 1$. Consider the following picture of $S^1$:

We are to understand that every point is supposed to be taken to its alike, from left to right. (E.g., the leftmost should be taken to the right one.) If that were the case, by considering the restriction of the alledged $F$ to the complement of the other points, i.e. $$F: S^1 \backslash \{p_1,\cdots,\widehat{p_i},\cdots p_n,q_1, \cdots, \widehat{q_i}, \cdots, q_n\} \to S^1 \backslash \{q_1, \cdots, \widehat{q_i}, \cdots, q_n\}, $$ it follows by connectedness that the orange arc should be taken inside the red arc. Likewise, the light-blue arc should be taken to the blue one, but this is a contradiction.

Another example which is closely related but perhaps more explicit is when $M=\mathbb{R}$. Pick $p_1<p_2<p_3, q_1<q_3<q_2$. By the intermediate value theorem, we would have some $x$ between $p_1$ and $p_2$ such that $F(x)=q_3$. But since $F(p_3)=q_3$, $F$ can't be injective.