# Why is $(D^1,S^0)$ not homotopy equivalent to $(\mathbb{R},\mathbb{R}\backslash\{0\})$ in TOP(2)?

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)?

• A continuous map $\mathbb{R}\to D^1$ that maps $\mathbb{R}\setminus \{0\}$ into $S^0$, what can you say about its image? Jan 11 '18 at 22:05
As discussed in the comments, the only continuous map of pairs $$(\mathbb{R}, \mathbb{R} \setminus 0) \to (D^1,S^0)$$ are constants. Such a map cannot be a homotopy equivalence since it misses a path component of $$S^0$$.