Let $M$ be an $n$-dimensional topological closed manifold. Suppose $K$ is a compact subset of $M$ which is contractible in the sense that there exists a continuous map $F:K \times [0,1] \to M$ with $F(\cdot,0)=id$ and $F(\cdot, 1)=q \in M$.


Can we find an open neighborhood $U$ of $K$ such that $U$ is homeomorphic to $\mathbb R^n$?



Contractible set in a manifold
This is not true. Let K be one component (it doesn't matter which one) of the [Whitehead link](https://en.wikipedia.org/wiki/Whitehead_link), which has two components. Then K is contractible in the complement M of the other component. But K is not contained in a 3-ball in M. This can be seen in many ways; for instance if K were contained in a 3-ball, then each of its lifts to the universal cover of M would have trivial linking numbers. But you can readily draw the picture of the cover and see that some of those linking numbers are $\pm 1$.


