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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!

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    $\begingroup$ A continuous map $\mathbb{R}\to D^1$ that maps $\mathbb{R}\setminus \{0\}$ into $S^0$, what can you say about its image? $\endgroup$ Jan 11 '18 at 22:05
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    $\begingroup$ The map has to be constant! Thanks! $\endgroup$
    – Joo
    Jan 11 '18 at 23:06
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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 $S^0$.

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