A Homotopy equivalence between two sets($A$ and $S^1$)

Question

I have to show the set $A=\{(x,y,z): x^2+y^2>z^2\}$ is homotopy equivalence to the $S^1=\{(x,y):x^2+y^2=1\}$. I have to find two continuos maps $f$ and $i$ such that $f\circ i$ is homotopy to the identity map from $S^1$ to $S^1$ and $i\circ f$ is homotopy to the identity map from $A$ to $A$. I have take the maps: \begin{align*} f\colon \quad A \quad &\longrightarrow \quad S^1, \\ \begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} &\longmapsto \begin{bmatrix} \frac{x}{||•||} \\ \frac{y}{||•||} \\ 0 \\ \end{bmatrix} \end{align*} where $||•|| $ denotes the norm of the vector$ \begin{bmatrix} x \\ y \\ 0 \\ \end{bmatrix}$ and \begin{align*} i\colon \quad S^1 \ \ &\longrightarrow \ \ \ A,\\ \begin{bmatrix} x \\ y \\ 0 \\ \end{bmatrix}&\longmapsto \begin{bmatrix} x \\ y \\ 0 \\ \end{bmatrix}. \end{align*}

But these maps aren't good (or not?) How can I find the homotopy?

Answer 1

$f(x,y,z)=({x\over{\sqrt{x^2+y^2}}},{y\over{\sqrt{x^2+y^2}}},0)$ $i(x,y,0)=(x,y,0)$.

$f(i(x,y,0)=(x,y,0)$, $i(f(x,y,0))=({x\over{\sqrt{x^2+y^2}}},{y\over{\sqrt{x^2+y^2}}},0)$

$h_t(x,y,z)=t(x,y,z)+(1-t)({x\over{\sqrt{x^2+y^2}}},{y\over{\sqrt{x^2+y^2}}},0)$