I saw a video (Who cares about topology) that explained the inscribed square problem. The problem say's that you have a simple closed loop in the plane, prove that there's at least one square such that all vertices are in the loop.
But I didn't understand a part of it.
In the video they say that if you take an ordered pair of points that are in the loop, you can map it to a specific point in a torus.
Then he show's that the same can be say for unordered pairs, but in a Möbius strip.
But, by that logic, Isn't possible to just glue two mobius strips (A and B) by their side, such that the pair of points (a,b) is in A, and (b,a) is on B.
What's the problem with that?
Why a torus and not a Klein bottle?
A simple closed curve is homeomorphic to $S^1$ (Circle).
So, for the pair of points $(a, b)$ it's possible to draw a circle, where a point in this circle corresponds to $a$ (or a point to the curve), and for each point in this circle, there's another circle that represents the points of $b$ (The same curve).
This mean that you have a shape that's $S^1$x$S^1$, and that shape is the torus.
Again, to see why the torus, but more visually.
So, you chose first a point from the purple circle ($a$), and then another point from red circle ($b$), thinking that both of those circles are the same than the simple closed curve.