Motivation for quotient topology

This page will discuss informally the motivation for the Quotient topology

Quotients (for example $X/\sim$ should already mean "gluing" or "associating" things together (like with an equivalence relation, $\sim$)

We will use this to talk about the topology of the torus $T$, from the real plane $\mathbb{R}^2$

Convincing yourself the torus is a quotient map

Imagine you're playing one of those old arcade games with a square, and if you go out of the top of the square you magically appear at the bottom, and if you go off the left side, you appear at the right, this is a torus. If you're on a doughnut and go "up" you'll go around and end up (eventually) back where you started, same for going left (around the torus)

How is this "gluing", well you can take a square and glue as this diagram shows:

$$\require{AMScd} \begin{CD} \circ @>A>> \circ\\ @VBVV @VVBV\\ \circ @>>A> \circ \end{CD}$$
Glue the "A" and "B" edges along the specified arrow

After gluing $A\leftrightarrow A$ one clearly has a tube, by wrapping that tube around and gluing $B\leftrightarrow B$ - a torus results.

Gluing $A\leftrightarrow A$

This means associating the point $(x,1)$ with $(x,0)$ - this is clearly Equivalence relation territory, but what does that mean for the point $(x,1.5)$ say? Well this is $0.5$ up from $(x,1)\sim(x,0)$ so is also $0.5$ up from $(x,0)$ - thus we have $(x,1.5)\sim(x,0.5)$

Going forward with this logic, we see after gluing $A\leftrightarrow A$ that we have:

$$(x,y)\sim_A(x',y')\iff[(\exists m\in\mathbb{Z}:y=y'-m)\wedge x=x']$$

Gluing $B\leftrightarrow B$

Using the same logic as above, $$(x,y)\sim_B(x',y')\iff[(\exists m\in\mathbb{Z}:x=x'-m)\wedge y=y']$$

Gluing $B\leftrightarrow B$ AND $A\leftrightarrow A$

We must now combine:

• $$(x,y)\sim_A(x',y')\iff[(\exists m\in\mathbb{Z}:y=y'-m)\wedge x=x']$$
• $$(x,y)\sim_B(x',y')\iff[(\exists m\in\mathbb{Z}:x=x'-m)\wedge y=y']$$

Which is simply: $$(x,y)\sim(x',y')\iff\exists m,n\in\mathbb{Z}:[x=x'-m\wedge y=y'-n]$$

(We cannot simply do "and" because of the part of $\sim_A$ for example, we could use the fact we have an orthagonal basis, but that ruins the topology part, so just simply take this as "common sense")

Mapping from the plane to the torus

Lets use $p:\mathbb{R}^2\rightarrow\tfrac{\mathbb{R}^2}{\sim}$ for the projection map that'll project our point on the plane to a point in $\mathbb{R}^2/\sim$, we can give $p$ explicitly as $p:(x,y)\mapsto[(x,y)]$ where $[(x,y)]$ denotes an Equivalence class - as usual.

The larger the topology the more "information" we are carrying over. For example suppose we had:

• $$p:(\mathbb{R}^2,\mathcal{J})\rightarrow(\tfrac{\mathbb{R}^2}{\sim},\{\emptyset,\tfrac{\mathbb{R}^2}{\sim}\}$$ (where $\mathcal{J}$ is the usual topology on $\mathbb{R}^2$ - you know involving open balls and such, and the topology on $\tfrac{\mathbb{R}^2}{\sim}$ is the indiscreet topology)

Well this is continuous ( $$\forall U\in\{\emptyset,\tfrac{\mathbb{R}^2}{\sim}\}$$ we have $$p^{-1}(U)\in\mathcal{J}$$ as indeed $p^{-1}(\emptyset)=\emptyset\in\mathcal{J}$ and $p^{-1}(\tfrac{\mathbb{R}^2}{\sim})=\mathbb{R}^2\in\mathcal{J}$)

But it isn't very useful!

Largest topology

A very natural thing to say would be "okay, we'll say that $$U\subseteq\tfrac{\mathbb{R}^2}{\sim}$$ is open, if $p^{-1}(U)$ is open (in $\mathbb{R}^2$)"