Homotopy equivalence of S1 and R2-0 I want to show that $X=S^1=\{x^2+y^2=1|x,y\in \mathbb{R}\}$ and $Y=\mathbb{R}^2-\{0\}$ are homotopy equivalent. For this I have to find a function $f:X\rightarrow Y$ and a function $g:Y\rightarrow X$ such that $f\circ g$ is homotopic to the $id_Y$ and  $g\circ f$ is homotopic to the $id_X$.
The construction of $g$ is easy: $g(x,y)=\frac{1}{\sqrt{x^2+y^2}}\begin{pmatrix}x\\y\end{pmatrix}$.
But how do I construct $f$? There is a certain ambiguity 'how much I stretch the unit circle'?
Many thanks in advance
 A: $\mathbb{S}^{n} \simeq \mathbb{R}^{n+1} - \lbrace 0 \rbrace$
Indeed, consider the functions: $$f: \mathbb{R}^{n+1} - \lbrace 0 \rbrace \longrightarrow \mathbb{S}^{n}$$ and $$i : \mathbb{S}^{n} \longrightarrow \mathbb{R}^{n+1} - \lbrace 0 \rbrace$$ so defined: $f(x) = \dfrac{x}{||x||}$ and and $i$, the inclusion. 
1) $(f \circ i) = \text{id}_{\mathbb{S}^{n}}$
2) $i \circ f : \mathbb{R}^{n+1} - \lbrace 0 \rbrace \longrightarrow \mathbb{R}^{n+1} - \lbrace 0 \rbrace $ is homotopic the identity application of $\mathbb{R}^{n+1} - \lbrace 0 \rbrace$  by means of a linear homotopy, since every point $x \neq 0$ in  $\mathbb{R}^{n+1}$  can be joined to the $\dfrac{x}{||x||}$ by means of a line segment which does not contain the origin. More precisely: $$H : \mathbb{R}^{n+1} - \lbrace 0 \rbrace \times I \longrightarrow \mathbb{R}^{n+1} - \lbrace 0 \rbrace $$ is the above-mentioned linear homotopy, thus defined: $$H(x,t) = (1-t)x + tf(x)$$ note that:
3) $H(x,0) = x = \text{id}_{\mathbb{R}^{n+1} - \lbrace 0 \rbrace}$
4) $H(x,1) = f(x) = (i \circ f)(x)$
The result follows from the items $(1)$ and $(4)$.
