# Proof that $f$ is not a homotopy equivalence

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 $$D^{2}-\{0\}$$.

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.

• Thanks Huang Samuel. You are right. – user Jun 1 at 17:44