I am having trouble understanding the proof of the following statement:
Let $X = T^2$ be the 2-dimensional torus defined as $$T^2 := \mathbb{R}^2 / \sim $$ where $(x,y) \sim (x',y')$ iff $x-x', y-y' \in \mathbb{Z}$. Prove the following subset is not a submanifold: $$ Y = \{[(t^2,t^3)] \in X \hspace{3pt} |\hspace{3pt} t \in \mathbb{R}\} $$
The proof is as follows: Assume it is a submanifold, then theres an open neighbourhood $U$ of $[(0,0)]$ and a homeomorphism $f : U \cap Y \to V$ where $V \subset \mathbb{R}$ open. If $\epsilon > 0 $ is sufficiently small then $U \cap Y$ contains the set $W_0$ of points $[(t^2,t^3)]$ where $-\epsilon < t < \epsilon$.
By definition of $X$ there exists $\epsilon ' >0$ sufficiently small such that $U \cap Y$ contains the set $W_1$ of points $[(s^2-1, s^3-1)]$ where $1-\epsilon' < s < 1+\epsilon'$.
From here notice $W_0$ and $W_1$ are distinct connected subsets of $U \cap Y$ containing $[(0,0)] $ which gives the contradiction since there are no open subsets of $\mathbb{R}$ containing connected subsets meeting at just one point.
My questions:
- I don't understand why $U \cap Y$ contains $W_1$. I think its to do with the equivalence relation but I don't immediately see why this is true.
- I don't understand why $W_0$ and $W_1$ are distinct connected subsets of $U \cap Y$ containing $(0,0)$.
- What is the intuitive idea behind why this cannot be a submanifold?
Working with $T^2$ I always find confusing because of the equivalence relation. I get really confused trying to visualise it all!
I would really appreciate any guidance.
The essential idea here is that if it were a submanifold then it should just look like an interval in a neighbourhood around $0$. However it looks like the picture below near $0$.
If $Y$ were submanifold it should have just been an interval, however here it looks like two crossed lines.