Is there always a homotopy taking a point to another in a connected manifold? 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 $a=e^{i\theta_{a}}, b=e^{i\theta_b}$.

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.
