# Forming manifolds from intersecting lines in Euclidean space

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.

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.
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.