# Cone is a topological manifold

This question is quite trivial, so please remove if not suitable. I'm revising topological spaces and would like some verification on a simple example.

Consider the surface $S$ in $\mathbb{R}^2$ given by $S=\{(x,|x|) \}$ which describes one dimensional cone. I claim that this is a topological manifold.

We consider the subspace topology on $S$ inherited from the usual topology on $\mathbb{R}^2$, and the usual topology on $\mathbb{R}$. Define the projector $p:S\rightarrow \mathbb{R}$ given by $p(x,|x|)=x$ which is continuous. It is also bijective with inverse $f:\mathbb{R}\rightarrow S$ given by $f(x)=(x,|x|)$ which is also continuous.

This chart covers all of $S$, and $S$ is obviously Hausdorff since $\mathbb{R}^2$ is. Hence $S$ is a topological manifold.

Is this correct? I ask because I've seen a few people claim that this cone is not a topological manifold.

• Yes, it's all correct. – Moishe Kohan Aug 23 '17 at 18:22
• Concerning the last sentence, it could be a miscommunication. The cone is not a differentiable submanifold of $\mathbb{R}^2$, and some people may just say that the cone is not a manifold, meaning it's not differentiable [well, we can endow it with a differentiable structure, it just doesn't inherit one from the ambient space], not that it's not a topological manifold. If people speak explicitly of topological manifolds, that interpretation doesn't work, of course. Then probably when people say "cone" they mean the double cone ($x^2 = y^2$ here). – Daniel Fischer Aug 23 '17 at 18:46
• Right. And, please, provide links to the "few people claim" (I do not think these are some random people whom you've met in a coffee shop). – Moishe Kohan Aug 24 '17 at 4:04