# The idea on how to show that the cone is not a topological surface

I just started taking Topology classes and we've been introduced to the concept of a topological surface, that is by Definition:

Def: A topological surface $S$ is a Hausdorff set such that $\forall x \in S, \exists U \ni x$ open neighborhood such that $U$ is homeomorphic to a subset of $\mathbb{R}^2$

My Professor then gave us a few examples and one particular counter example that I still can't wrap my head around:

Example: The set $C= \lbrace (x,y,z) \in \mathbb{R}^3 \mid z^2 = x^2 + y^2 \rbrace$ is not a topological surface

When he asked the class I answered that the origin, i.e. the point $(0,0,0) \in C$ would cause trouble but I would not know on how to prove it. He affirmed that idea to me and said that I should think about removing the point and think about

Lemma: $f: X \to Y$ homeomorphism, $X$ connected $\implies Y$ connected

So that is what I did for the last couple of hours and I have confused myself utterly.

My problems:

Let me first start with the things that are clear to me:

1. If $U \subset \mathbb{R}^2$ is connected, then $U\setminus \lbrace c \rbrace$ for $c \in \mathbb{R}^2$ is still connected
2. If I take $U \ni (0,0,0)$ open neighborhood and consider $C \cap U \setminus \lbrace (0,0,0) \rbrace$ then this set is clearly disconnected, I did split up the cone locally around its origin in the upper half and lower half

Now to my problems:

• I cannot prove that $C \cap U\setminus \lbrace (0,0,0) \rbrace$ is not homeomorphic to a subset of $\mathbb{R}^2$ although it is intuitively very clear
• I don't see how to apply the Lemma
• Far worse: I don't see why I want to do anything of the above.

If I negate the statement of a topological surface I read it as: $$\exists x \in S, \forall U \ni x \text{ open neighborhood}: U \text{ is not homeomorphic to } \mathbb{R}^2$$ So my point of interest (i.e. $(0,0,0)$) should still be in the neighborhood $U$ I consider right? I am sure that I need to raise a contradiction with all the listed ideas above, but I don't see the connection to $C$ because I have removed informations from $C$.

Assume that $C$ is a topological surface.

It is our aim is now to find a contradiction.

Open sets $U\subseteq C$ and $V\subseteq\mathbb{R}^{2}$ exists together with a homeomorphism $\phi:U\rightarrow V$ and $\langle 0,0,0\rangle\in U$.

For $\langle a,b\rangle:=\phi(\langle 0,0,0\rangle)\in V$ some open ball $B\subset\mathbb{R}^{2}$ centered at $\langle a,b\rangle$ exist with $\langle a,b\rangle\in B\subseteq V$.

Let $W:=\phi^{-1}\left(B\right)$ and prescribe $\psi:W\rightarrow B$ by $w\mapsto\phi\left(w\right)$.

Then $\psi$ is a homeomorphism.

However, it sends the not connected set $W-\left\{ \langle 0,0,0\rangle\right\}$ to the connected set $B-\left\{ \langle a,b\rangle\right\}$.

A contradiction has been found. Actually I used the converse of the lemma (a homeomorphism sends not connected sets to not connected sets) wich is also true. This because homeomorphisms have inverses, and these inverses are also homeomorphisms.

• Thanks a lot for your effort, really well written! You brought clarity into this topic for me. Commented Sep 26, 2015 at 16:53
• After reviewing this I wonder why it's not possible to consider $$\phi: U - \lbrace (0,0,0) \rbrace \to V - \lbrace \phi(0,0,0)\rbrace$$ and then conclude the same, $U - \lbrace (0,0,0) \rbrace$ is not connected while $V- \lbrace \phi(0,0,0) \rbrace$ still is connected. In other words, why is the 'route' through $\Psi$ necessary? Commented Sep 26, 2015 at 17:53
• Then $V$ must be connected set on forehand. That is not guaranteed. To avoid $\psi$ you could use a broader lemma: If $f:X\rightarrow Y$ is a homeomorphism then connected subsets of $X$ correspond with connected subsets of $Y$ and vice versa. Commented Sep 26, 2015 at 20:31