Intuitive way of thinking about quotient topology

Question

I am having trouble visualising the quotient topology for some of the "wackier" spaces. For example, I can understand why a circle is [0,1]/0~1, but I have no idea what to think of if you take a disc and identify two points on its boundary. Help on this specific example and more generally would be appreciated!

EDIT: In particular, I am looking for how to imagine what a the space would look like (if it can embedded in $\mathbb{R}^3$) and how one would go about computing the fundamental group.

Answer 1

The quotient topology formalizes the intuitive notion of "gluing points together". We can make a partition of the topological space whose equivalence classes correspond to points we want to identify. When we make the equivalence relation, we keep only the open sets which respect this equivalence relation. These are the ones which are unions of equivalence classes.

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A Useful Universal Property

A useful way to think about it is through this fact, called a universal property: if there is a continuous map $f : X \rightarrow Y$ and an equivalence relation $\sim$ on $X$ such that $x \sim y$ implies $f(x) = f(y)$, then $f$ induces a continuous function $\tilde{f} : X/\sim \rightarrow Y$. There is also a continuous map $\pi : X \rightarrow X/\sim$ where $x \in X$ is sent to its equivalence class. Then $\tilde{f} \circ \pi = f$. We can write what we have like this:

This is helpful in practice, but how does one get a visual intuition for what it means to "identify points"? For this, we can work out some examples.

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Examples

Consider the original example, where $I / \sim \ \cong S^1$ ($\sim$ is the smallest equivalence relation such that $0 \sim 1$). Identify the circle $S^1$ with $\{ z \in \mathbb{C} : |z| = 1 \}$. Define $\phi : I \rightarrow S^1$ where $t \mapsto e^{2 \pi i t}$. Then $\phi (0) = \phi(1)$, so that, with the equivalence relation you wrote, $\phi$ factors through $I / \sim $ giving a morphism $\tilde{\phi} : I / \sim\ \rightarrow S^1$. We can then show that $\phi$ is an isomorphism using this

Another example: take the square $I^2$ and identify $(0, t)$ with $(1, t)$ and $(1, s)$ with $(0, s)$. We can show that this is isomorphic to $S^1 \times S^1$ (the torus). Intuitively this is gluing together points like below:

Let's make this formal. Define a map $\phi : I^2 \rightarrow S^2$ where $(t, s)$ is sent to $(e^{2 \pi i t} , e^{2 \pi i s})$. It's not hard to see that $\phi((1, t)) = \phi((0, t))$ and $\phi((t, 1)) =\phi((t, 0))$. Using the universal property above, $\phi$ factors through a map $\tilde{\phi} : I^2 / \sim \ \rightarrow S^1 \times S^1$. This is a continuous bijection, so using this we see that it's a homeomorphism.

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Calculating Fundamental Groups

You asked about how one calculates the fundamental group. In practice, calculating the fundamental group of spaces can be done using a standard procedure using CW-complexes. This is the approach one would use to find the fundamental group of a disc with two points glued together. There is also something called the van Kampen Theorem:

Theorem: take a path connected space $X$ with base point $x$. Let $\mathcal{U}$ be an open cover of $X$ closed under finite intersections and each containing $x$. Then, if we think of $\mathcal{U}$ as a category whose morphisms are inclusions of subsets, $\pi_1$ induces a functor $\Phi : \mathcal{U} \rightarrow \text{Grp}$. We have $\pi_1(X, x) \cong \text{colimit } \Phi$.

If you know the fundamental group of each element of the open cover, this can be used to calculate the fundamental group of $X$. For instance, suppose we would like to calculate the fundamental group of $S^2$. Viewing $S^2$ as a subspace of $\mathbb{R}^3$, we can define open sets

$U = \{ (x, y, z) \in S^2 : z > -1/2 \}$

$V = \{ (x, y, z) \in S^2 : z < 1/2 \}$

$W = \{ (x, y, z) \in S^2 : -1/2 < z < 1/2 \}$

$\{ U, V, W \}$ is an open cover of $X$. We can see from the van Kampen theorem that the fundamental group of $S^2$ is going to be isomorphic to a pushout like this:

But such a pushout must be $0$ (the trivial group). So $\pi_1 ( S^2) = 0$ (it didn't matter what base point we choose).

I hope this helped you think about quotient spaces and fundamental groups!

Answer 2

One mental algorithm that sometimes helped me visualize constructions of quotient topologies is as follows:

1. Imagine the original space as made out of rubber and floating in vacuum and zero gravity, spread out nicely, if possible. For example, $S^2$ in $\mathbb{R}^3$ would appear as inflated, although with zero pressure outside and inside.
2. Now imagine that all the points in the equivalence class that you are modding by are drawn to each other (each point to all other points) by a really strong force, to the point where they collapse into a single point.

It's easy to imagine what happens to $S^2$ if you identify the north and the south poles, for example. But often this "algorithm" is hard to use when the identification has no intuitive presentations (e.g. $\mathbb{R}/\mathbb{Q}$).