Topological Quotients: Understanding $X/\sim$ and $X/Y$ with Insights into the disk Structure.

Question

I could use some assistance in clarifying a concept. In topology, when we have a space denoted as $X$, we can create a quotient space (a space of equivalence classes) denoted as $X/\sim$, where $\sim$ is an equivalence relation defined on $X$. However, I'm curious about the meaning of $X/Y$ when both $X$ and $Y$ are topological spaces. Specifically, I'd like to understand why the quotient $S^1 \times [0,1]/S^1 \times \{1\}$ is referred to as a disk $D^2$.

Answer 1

As @SassatelliGiulio already notes in the comments, $X / Y$ in the context of topological spaces where $Y \subseteq X$ is a subspace means the quotient $X / {{\sim_Y}}$ where $a \sim_Y b$ if $a, b \in Y$ (or $a = b$). In other words, $X / Y$ is obtained from $X$ by taking all the points in $Y$ and identifying them to be a single point. We also say that $X / Y$ is obtained from $X$ by collapsing $Y$.

For your concrete example, $S^1 \times [0, 1]$ is a cylinder, and collapsing the "top" (or bottom, depending on which way you draw it) boundary circle $S^1 \times \{1\}$ gives you a cone, which is homeomorphic to $D^2$. More formally, the map $f\colon S^1 \times [0, 1] \to D^2$ given by $f(x, t) = (1 - t)x$ is continuous and $f(x, 1) = 0$ for all $x \in S^1$, so by the universal property of quotient spaces it factors over a map $S^1 \times [0, 1] / S^1 \times \{1\} \to D^2$ (note for this that a map $g\colon X \to Z$ respects $\sim_Y$ iff $g|_Y$ is constant) which is in fact a homeomorphism (I leave checking the details of this to you :)

Answer 2

As mentioned in the comments, the quotient of a space $X$ by a subspace $Y$ is the space whose points are equivalence classes such that all of $Y$ is a single class (a point in the quotient), and every point in $X \setminus Y$ is a singleton class. So, we often speak of collapsing $Y$ to a point in $X$, thinking of the quotient map $X \twoheadrightarrow X/Y$.

In your example, $X = S^1 \times [0, 1]$ is the cylinder over the circle, and the subspace $Y = S^1 \times \{1\}$ is the circle at one end of the cylinder. What happens when you pinch all of that circle to a single point under the quotient $q$?

You get a cone. Now, you have to convince yourself that this cone is homeomorphic to the disk $D^2$. Picture pushing the cone point down to the plane of the bottom circular boundary, effectively smashing it into the plane. Can you write down this map $h: X/Y \to D^2$ explicitly?