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Let $M$ be a connected topological $n$-manifold. Note this implies $M$ is path-connected.

Let $a,b \in M$. Must there always exist a continuous $H:M \times [0,1] \to M$ such that $H(m,0)=m$ for all $m \in M$, and $H(a,1)=b$?

We can note that this is true for $\mathbb{R}^n$, by considering $(x,t) \mapsto x+t(b-a)$, and it is also true for $S^1$ by considering $(e^{i\theta}, t)\mapsto e^{i(\theta+t(\theta_b-\theta_a))}$ where $a=e^{i\theta_{a}}, b=e^{i\theta_b}$.

MathematicsStudent1122
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    The relation "$a$ can be reached from $b$ by a self-homotopy of $M$" is an equivalence relation. Check that each equivalence class is open, hence closed (since its complement is the union of the other equivalence classes), hence the whole manifold $M$ (since $M$ is connected). – Andreas Blass Nov 24 '20 at 19:03
  • @AndreasBlass That is a very slick answer! Thank you! – MathematicsStudent1122 Nov 24 '20 at 19:13

3 Answers3

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If $M$ is a topological manifold and $a\in M$ is any point, then the inclusion $\{a\}\hookrightarrow M$ is a cofibration (i.e. has the homotopy extension property). Take a path $\gamma:I\rightarrow M$ from $a$ to $b$ and use the HEP to get a homotopy $H:M\times I\rightarrow M$ with $H_0=id_M$ and $H_1(a)=\gamma(1)=b$.

Tyrone
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Let’s prove something stronger: for each such manifold $M$ (without boundary), the path-connected component of the identity in the group of homeomorphisms of $M$ acts transitively on $M$.

Since it’s a continuous group action and $M$ is connected, it’s enough to show that any orbit is open. By considering homotopies that are the identity outside a ball, we can assume $M=B^n$, $a,b$ being interior points, and require that the homotopy must be the identity on the boundary.

But in this case, you can consider the flow of a compactly supported vector field in the right direction.

Aphelli
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For manifolds with boundary: no, you can't take boundary points to non-boundary points.

For manifolds without boundary: yes. For 1-manifolds, this is easy. For $n > 1$, take a neighbourhood of a (shortest) path between your points. After removing problems (by taking a subset), that's homeomorphic to a disk, with both points in the interior. Take any homeomorphism of the disk that interchanges the points while fixing the boundary, and pass it through the homeomorphism to the neighbourhood of the path.

user3482749
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    Your first statement is only true if $H$ is required to be an isotopy through homeomorphisms. It's easy to put a collar on the boundary and deformation retract it into the interior. – Tyrone Nov 24 '20 at 19:14