retract by deformation I'm learning about deformation retract and have this problem as an exercise:
let  :
$$V:=\left\{(x, y, z) \in \mathbb{R}^{3} / x^{2}+y^{2} \neq 0\right\}$$
Show that $W:=\left\{(x, y, 0) / x^{2}+y^{2}=1\right\}$ is a retract by deformation of V
If you can help me I will thankful to you.
 A: It might be helpful to visualize it: $V$ is the $\Bbb R^3$ without the $z$-axis and $W$ is the circle with radius $1$ and center $(0,0,0)$ inside the $x$-$y$-plane. Define the map $r:V\to W$ by first "flattening", i.e setting the $z$-coordinate $=0$, and then contracting everything to the circle along rays: $$r(x,y,z)=\frac{1}{\sqrt{x^2+y^2}}(x,y,0)$$
Show that this is a retraction. Now construct a homotopy between the identity on $V$ and $r$. You can do that in the simplest possible way, i.e. connecting two points $x$ and $r(x)$ via a straight line (why?).
Edit: A homotopy between $\operatorname{id}_V$ and $r$ (or more formally between $\operatorname{id}_V$ and $j\circ r$ where $j:W\hookrightarrow V$ is the inclusion) can be defined as follows:\begin{align*}
H:V\times[0,1]&\to V\\
(x,t)&\mapsto (1-t)x+t\cdot r(x)
\end{align*}
I.e. $H(x,t)$ is the point on the straight line between $x$ and $r(x)$ linearly parameterized by $t$. Now check what you get for $t=0,1$ and that the homotopy is indeed inside $V$.
