Trouble understanding the torus as a quotient space of the unit square.

Question

Let $X=[0,1]\times [0,1]$ be the unit square in $\mathbb R^2$ and define a partition $X^*$ of $X$ by

• All one-point sets $\{(x,y)\}, \ 0<x<1, 0<y<1$
• All two-point sets $\{(x,0), (x,1)\}, \ 0< x< 1$
• All two-point sets $\{(0,y),(1,y)\}, \ 0<y<1$
• The four-point set $\{(0,0),(0,1),(1,0),(1,1)\}$.

Let $p:X\to X^*$ be the surjective map that carries each point of $X$ to the element of $X^*$ containing it, then in the quotient topology induced by $p$, the space $X^*$ is called a quotient space of $X$.

The intuition here is that the edges of the unit square are "glued together" to form the torus. I am having trouble understanding this example rigorously. For a given $(x,y)\in X$, what exactly is $p(x,y)$? What exactly are the elements of $X^*$? Would I be correct in saying that $X^*$ has four elements, and for example $$ p(1/2,1/2) = \{(x,y): 0<x<1,0<y<1\}\quad ? $$ If this is the case, how can I see $X^*$ as the torus, which (presumably) is a subset of $\mathbb R^3$? I am sorry if my exact question is not clear, because this example is very confusing to me. Can anyone lend some insight into better understanding this quotient space?

Answer 1

Consider the square $[0,1]\times [0,1]$, when you say identify $(0,t)\sim(1,t)$ it means, rolled up a piece of paper of size $[0,1]\times [0,1]$ in a way that, upper edge overlap with lower edge, and this gives you a cylinder with horizontal axis. Now we also assume that, $(t,0)\sim (t,1)$ on the square $[0,1]\times [0,1]$. But the points $(t,0),(s,1)$ with $0\leq t,s\leq 1$ now are on the two circular edges of cylinder. So to identify $(t,0)\sim(t,1)$ in square is now same as identifying the point $c_{(t,0)}$ on the left hand side of cylinder to the $c_{(t,1)}$ on the right hand side of the cylinder. So you have to bent the cylinder such that, $c_{(t,0)}$ coincide with $c_{(t,1)}$ i.e. join the left side of the cylinder to the right side of the cylinder without any perturbation or twist. This gives you torus.

Now, the edges of the square $[0,1]\times [0,1]$ after identification gives you a shape of figure-eight or $\Bbb S^1\lor \Bbb S^1$ on the torus $T$. While the interior points of $[0,1]\times [0,1]$ will associate to the points of $T\backslash (\Bbb S^1\lor\Bbb S^1)$ in one-to-one and onto fashion. Here $\Bbb S^1\lor\Bbb S^1$ means the joining of two circles exactly at one point.

While constructing torus from $[0,1]\times[0,1]$ if you consider the edges of this square, then as a first step of the identification $(t,0)\sim (t,1)$ gives you a line segment with two circles hanging on two ends of the segments. Next, as a second step of the identification $c_{(0,t)}\sim c_{(1,t)}$ ends up with identifying two ends of the previous line segment as well as identifying the two circles which are hanging two ends of the line segment, i.e., a figure with a shape like two circles joined or wedge exactly at one point, and this wedge point is nothing but the equivalence class $\{(0,0),(0,1),(1,0),(1,1)\}\in T$.

Answer 2

The elements of X* are any element of any of those four bullets you posted. This effectively leaves the interior untouched, but collapses opposite points along each edge into the same element, gluing them together. Likewise, the corners are all collapsed together into one point.

This torus can be embedded in $\mathbb R^3$, sure, but it doesn't have to be. You can think of it as its own space. Just like $\mathbb R^3$ can be embedded in $\mathbb R^4$, but you don't have to consider $\mathbb R^4$ when analyzing $\mathbb R^3$.