How to prove a cone is homeomorphic to $\mathbb{R}^2$

Question

I wish to prove the infinite cone $T$ (where $T$ is the upper half i.e. $z \geq0$ )is homeomorphic to $\mathbb{R^2}$.

Note here that I am considering $T$ as a subset of $\mathbb{R}^3$ with the subspace topology and the standard Euclidean topology on $\mathbb{R^3}$ and $\mathbb{R^2}$ with the standard topology as well.

I have constructed a map $f:T \to \mathbb{R^2}$ as $f((x,y,z))=(x,y)$ (Basically we just project each point vertically down onto the $xy$ plane). And the inverse map $f^{-1}: \mathbb{R^2} \to T$ by $f^{-1}((x,y))=(x,y,(x^2+y^2)^{1/2})$ (this map just projects up until we hit the surface).

Clearly $f$ bijective and I can show $f$ is continuous. But I don't know how to show $f^{-1}$ is continuous (which is the last step I need to show the homeomorpism).

I tried to take an open set in $T$ and show that its preimage was open but I couldn't do it.

Could anyone show me how to proceed?

Thanks!

Answer 1

We are going to use:

Theorem. An application $h:X\to Y$ is continuous if, and only if, $h:X\to h(X)$ is continous, considering $h(X)$ as a topological subspace of $Y$.

Proof: note that for any subset $B\in Y$, $$h^{-1}(B)=\{x\in X: h(x)\in B\}=h^{-1}(B\cap h(X))$$

so if $G$ is an open set in $h(X)$, then it will be $G=h(X)\cap \tilde G$, with $\tilde G$ an open set in $Y$. Now: $$h^{-1}(G)=h^{-1}(G\cap h(X))=h^{-1}(\tilde G)$$

so $h^{-1}(G)$ is an open set in $X$ if, and only if, $h^{-1}(\tilde G)$ is an open set in $X$. That means $h:X\to Y$ is continuous if and only if $h:X\to h(X)$ is continuous.

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Now, Let us call $g=f^{-1}$. The application $g:\mathbb R^2 \to T, g(x,y)=(x,y,\sqrt{x^2+y^2})$ is continuous if and only if the following application $$g:\mathbb R^2 \to \mathbb R^3, g(x,y)=(x,y, \sqrt{x^2+y^2})$$ is continuous. Now, you can use the universal property of the product space to easily verify that $g$ is continuous.

Answer 2

Your $f^{-1}$ is a composition of lots of well-known continuous functions. For example, $\mathbb R^2 \to \mathbb R^2$ mapping $(x,y) \mapsto (x^2, y^2)$ is continuous - you can show this from the fact that $\mathbb R \to \mathbb R$ mapping $x \mapsto x^2$ is continuous, and from the fact that square-shaped open sets form a basis for the standard topology on $\mathbb R^2$. The function $\mathbb R^2 \to \mathbb R$ mapping $(x,y) \mapsto x + y$ is also continuous. And so is $\mathbb R \to \mathbb R$, \mapping $x \mapsto \sqrt{x}$. Then use the fact that compositions of continuous functions are continuous.