Forming manifolds from intersecting lines in Euclidean space

Question

The textbook I'm reading introduces manifolds and gives some examples. Example 6 is, "Let $M=\{(x,y)\in\mathbb{R}^2:|x|=|y|\}$. This cannot be a manifold, since every neighborhood of $(0,0)$ decomposes $M$ without that point into four rather than two components, and consequently cannot be mapped homeomorphically onto an open interval."

Could someone explain to me why the decomposition into 4 parts results in the inability to map it to an open interval? I see that it seems impossible to map this set homeomorphically onto $\mathbb{R}$, but I am confused on the reasoning stated previously as to why the decomposition is a reason it cannot be mapped to $\mathbb{R}$.

I am guessing if one had the set $M=\{(x,y)\in\mathbb{R}^2:x=y\}$, that one could map this set homeomorphically onto $\mathbb{R}$ by simply mapping the point to its $x$-coordinate on the real line?

If it's of any interest, the textbook is Classical Mathematical Physics by Walter Thirring.

Answer 1

Intuitively, the negative reals a little bit less than 0 have to be mapped to one component, and the positive reals a little bit greater than 0 have to be mapped to one component. So we can't map things homeomorphically.

It's easy to translate this intuition into a proof. Let $E = \{(x,y) \in \mathbb{R}^2 : |x| = |y|\}$ and $f: E \to U$ a homeomorphism, where $U \subseteq \mathbb{R}$ is an open interval. WLOG $f((0,0)) = 0$. Then $U \cap (-\infty,0)$ is a connected set, so $f^{-1}(U\cap (-\infty,0))$ is connected and omits $(0,0)$, and thus it lies in one of the four connected components. Similarly with $U \cap (0,\infty)$. Therefore, $f^{-1}$ is not surjective onto $E$ and so, as expected, $f$ is not a bijection, contradiction.

Answer 2

Suppose that $M$ is a $1$-manifold. Each $x\in M$ has a neighbourhood $U$ in $M$ which is homeomorphic to the open interval $(-1,1)$ with $x\in U$ corresponding to $0$. Then $U\setminus\{x\}$ is homeomorphic to $(-1,0)\cup(0,1)$ which has two connected components.

But if $M$ is two intersecting lines, $x$ is their intersection, and $U$ is a neighbourhood of $x$, then $U\setminus \{x\}$ has at least four connected components, so $M$ cannot be a $1$-manifold. The reason for this is that $M\setminus \{x\}$ has four connected components and $U$ intersects each of these components non-trivially.