Union of the two coordinate axes in $\mathbb{R^2}$ - From the book '' Differential Topology '' of Guillemin and Pollack

Question

I have tried to solve the problem : Prove that the union of the two coordinate axes in $\mathbb{R^2}$ is not a manifold.

Let $X = \{(x,y) \in \mathbb{R^2} : x=0~or~ y=0\}$ be the union of the two coordinate axes in $\mathbb{R^2}$.

What happens to a neighborhood of $0$ when $0$ is removed?

I would like to finalize this question - issue by myself. I just need an explanation. Is anyone is able to help me?

Answer 1

How many connected components must any neighborhood of $0$ in $X$ have? How many connected components does an open interval in $\mathbb{R}$ with a point removed have? So can there be a homeomorphism between a neighborhood of $0$ in $X$ and an open interval in $\mathbb{R}$?

Answer 2

I can't lead with Micheal Joyce response. Someone could explain his comment in detail or suggest me another easier to use. The number of connected components seems interesting, but I can't use that information. How do I '' count '' the latter.

Answer 3

I found the problematic.

Removal of the point $0$ breaks the set into four connected components, something that cannot happen to any manifold. An alternative argument is that if $M \subset \mathbb{R}^2$ is any manifold and $x \in M$ then by the inverse function theorem, in some neighbourhood of $x \in M$, either the projection to the $x_1$-axis, or the projection to the $x_2$-axis is a local diffeomorphism, and in particular is one-to-one. Neither is one-to-one in this case.