If $A \subset X$ is contractible, then is $X \setminus A$ homotopy equivalent to $X \setminus a_0$?

Question

As in the title: let $A \subset X$ be contractible and let $a_0 \in A$. Are $X \setminus A$ and $X \setminus a_0$ homotopy equivalent? I tried to prove that it is but honestly i have no idea what a homotopy inverse to the inclusion $X \setminus A \rightarrow X \setminus a_0$ would look like. This is pretty difficult for me to visualize so any intuitive proof/counterexample would be greatly appreciated, or any general advice on looking for a homotopy inverse.

Answer 1

Nope! As a counterexample, take $X = \Bbb R^2$ and $A$ to be the $x$-axis. $A$ is contractible. However, $X \setminus \{0\}$ is path-connected, while $X \setminus A$ is not.