Let $X$ be a topological space and $A$ and $B$ are subspaces of $X$. Suppose that $A$ is contractible. I know that taking the quotient does not affect the homotopy type, that is $X/A$ is homotopy equivalent to $X$.
Do we have also a general result when taking the complement, I mean do we have that $X-A$ is homotopy equivalent to $X-*$ where $*\in A\subset X$ is the base point of $X$ ? If the answer is no then what if the subspace $A$ deformation retracts onto $*$ and not only having the homotopy type of $*$ ? Thank you for your help!
Consider $\mathbb{R} \setminus \mathbb{R} = \varnothing$, which does not have the same homotopy type as $\mathbb{R} \setminus \{0\} \sim S^0$ even though $\mathbb{R}$ deformation retracts onto a point. For more complicated examples (because this one could look "pathological"), you can remove a half-line from $\mathbb{R}$ and get a contractible space (not homotopy equivalent to $S^0$). Or you can remove a point from $\mathbb{R}^2$, or a line, or the union of two lines, etc., and you'll get differing homotopy types every time.