The cone is homeomorphic to the plane Prove that the cone $$\{(x,y,z):x^2+y^2=z^2, z\ge 0\}$$is homeomorphic to the plane
I know that an homeomorphism is a bijective fuction whose inverse is also continue. But I cannot imagine a correspondence between them.
Thanks for your help.
 A: I would like to remind you a definition and a theorem of interest:

Definition: if $f : X \to Y$ is any function and $Y \subset X$ is any subset, the restriction of $f$ to $Y$ is $$f|_{y}: Y \to Z \\ f|_{y}(x)=f(x)$$

Building up on this Definition we end up with:

Theorem 1: if $X,Z$ are topological spaces and $Y \subset Z$ has a subspace topology and if $f : X \to Z $ is continuous then $f|_{y} : Y \to Z$ is also continuous

If you want to prove this let me give a sketch. for Theorem 1 consider the map $$ i: Y \to X \\ x \to x$$ where $Y \subset X$. Then consider $f \circ i$ ($f$ composition $i$).
Now let us denote the cone by $C$.
Consider the map $$F : R^3 \to R^2 \\ (u,v,z) \to (u+v,u+v-z)$$
Now $F$ is continuous since it's coordinates are. $F$ is also a bijection. $F^{-1}$ exists. $F^{-1}$ is also continuous. Now restrict $F$ to the cone:
$$ G := F|_{C} : C \to R^2 $$
$G$ is continuous follows from the theorem. Now restrict $F^{-1}$ on $C$
$$H := F^{-1}|_C : R^2 \to C$$
By theorem $H$ is continuous. So Indeed $C$ and $R^2$ are homeomorphic to each other.
