I found the statement from the title in some introductory lecture notes on algebraic topology:
A homotopy in TOP(2) is defined as follows:
Let $f, g: (X,A)\rightarrow (Y,B)$ be two continuous maps. A homotopy between them is a continuous map $$H: (X\times [0,1],A\times [0,1])\rightarrow (Y,B)$$ which satisfies $$ H(\cdot,0)=f\quad \text{and}\quad H(\cdot, 1)=g$$ Two spaces $(X,A)$ and $(Y,B)$ are said to be homotopy equivalent, iff by definition there exist $f: (X,A)\rightarrow (Y,B)$ and $g:(Y,B)\rightarrow (X,A)$ such that $g\circ f$ and $f\circ g$ are homotopic to $id_{(X,A)} $ respectively $id_{(Y,B)}$.
In the proof of the statement in the title it only says $\overline{\mathbb{R}\backslash \{0\}}=\mathbb{R}$ and $S^0$ is not connected.
Those statements are obvious to me. I just don't see in what way this can be applied to see that there is no homotopy equivalence in TOP(2). Why is $(D^0,S^0)$ not homotopy equivalent to $(\mathbb{R},\mathbb{R}\backslash{0})$ in TOP(2)?
Thanks in advance!