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Let $f:(D^2, S^1) \rightarrow (D^2, D^2-\{0\})$, that is, $f:D^{2} \rightarrow D^2$ is continuous and $f(S^1) \subset D^{2}-\{0\}$. Show that $f$ is not homotopy equivalence.

I would like to show this exercise by contradiction and I should use that $0$ is a limit point of $D^{2}-\{0\}.$

Suppose that $f$ is a homotopy equivalence. Therefore there is $g:(D^2, D^{2}-\{0\}) \rightarrow (D^2, S^1)$ such that

\begin{eqnarray} f\circ g &\simeq& Id_{(D^2, D^2-\{0\})}\\ g \circ f &\simeq & Id_{(D^2, S^1)}. \end{eqnarray} Therefore there is $F:(D^2 \times I, (D^{2}-\{0\}) \times I) \longrightarrow (D^2, D^2-\{0\})$ and $G:(D^2 \times I, S^1 \times I) \rightarrow (D^2, S^1)$ such that \begin{eqnarray} \begin{cases} F(x,0)=(f\circ g)(x)\\ F(x,1)=Id_{D^2}(x) \end{cases} \end{eqnarray}

e

\begin{eqnarray} \begin{cases} G(x,0)=(g \circ f)(x)\\ G(x,1)=Id_{D^2}(x) \end{cases} \end{eqnarray}

$\forall \, x \in D^{2}.$

In this point, I don't know how to use that $0$ is a limit point of $D^{2}-\{0\}$.

Thanks in advanced.

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The map $g:(D^2,D^2\smallsetminus\{0\})\to(D^2,S^1)$ will map $0$ into $S^1$ by continuity of $g$.

(for example, using your hint, we may consider choosing a sequence converging to $0$, then the image of the sequence is also convergent, and converges into $S^1$ by compactness of $S^1$.)

Therefore $g$ maps the entire $D^2$ into $S^1$.

Since $g$ is the homotopic inverse of $f$, we obtained a homotopic equivalence between $D^2$ and $S^1$,

(To be more precise, consider the factored map $g':D^2\to S^1$ obtained from $g$ and the restriction $f':S^1\to D^2$ of $f$ ; these maps will be mutually homotopic inverses.)

which will be a homotopic equivalence between a contractible space and a non-contractible space.

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    $\begingroup$ Thanks Huang Samuel. You are right. $\endgroup$ – user Jun 1 at 17:44

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