# Analogy between quotient groups and quotient topology

I am trying to see "analogy" between the two concepts in the title, although I am familiar with independent definitions of them.

(1) Let $G$ be a group, $N$ a normal subgroup. Define an equivalence relation on $G$ by $g_1\sim g_2$ if $g_1g_2^{-1}\in N$. Let $[g_1]$ denote equivalence class of $g$, and $G/N$ the set of equivalence classes. Then we have a canonical map $\varphi\colon G\rightarrow G/N$, where domain is group, and codomain is "set" right now.

Then the quotient structure on $G/N$ is a "group structure" such that the map $\varphi$ becomes a homomorphism. Such a structure is unique. We call $G/N$ quotient group.

(2) Let $X$ be a topological space, $\sim$ an equivalence relation on $X$, and $[x]$ denote equivalence class. Let $X/{\sim}$ denote the set of equivalence classes. Then there is canonical map $\varphi\colon X\rightarrow X/{\sim}$, where domain is topological space and codomain is just set right now.

The quotient topology on $X/{\sim}$ is a topology such that the map $\varphi$ is continuous.

But, I was feeling that there could be more than one topological structures on $X/\sim$ which make the canonical map continuous. For example, take nice example of quotient topology on one hand, and structure of indiscrete topology on $X/{\sim}$ on the other hand.

Question: What is exact definition of quotient topology which makes it unique in some sense? (compare with quotient structure in (1))

• Re: "The quotient topology on $X/{\sim}$ is a topology such that the map $\varphi$ is continuous." Not only that. It is the finest topology such that $\varphi$ is continuous. See also Wikipedia article on final topology. – Martin Sleziak Sep 2 '16 at 5:28
• I saw wiki, there, they said in final topology on $X$ w.r.t. a family of functions into $X$; should it be From $X$ instead of into $X$? – p Groups Sep 2 '16 at 5:56
• No, it should be into $X$. The wiki has a different notation than you post, $X$ in the Wikipedia article corresponds to your $X/{\sim}$. (In your case, the family of functions consists on of the the single function $\varphi$.) – Martin Sleziak Sep 2 '16 at 6:02

The quotient topology is the unique topology on the set $X/{\sim}$ of equivalence classes such that for any topological space $Y$ and any map $f:X/{\sim}\to Y$, $f$ is continuous iff $\varphi\circ f$ is continuous. (Note that in the case where you take $Y=X/{\sim}$ and $f$ to be the identity map, this says in particular that $\varphi$ itself is continuous.) The analogous characterization also works for the quotient group structure, and pretty much any other quotient structure you'll ever run into.

Another way to phrase this in the case of topological spaces is that the quotient topology is the finest topology that makes $\varphi$ continuous. Indeed, for any topology that makes $\varphi$ continuous, $\varphi\circ f$ will be continuous whenever $f$ is continuous. So assuming $\varphi$ is continuous, the condition above is just that $f$ is continuous whenever $\varphi\circ f$ is: that is, there are as many continuous maps out of $X/{\sim}$ as possible (given the condition that $\varphi$ is continuous). More continuous maps out of a space means a finer topology, so this is saying it's the finest topology for which $\varphi$ is continuous.

• I am understanding your points; it is becoming quite clear. But I will work out details, since this is first time I am looking it so closely. Thanks for the insight into the definitions and analogy. – p Groups Sep 2 '16 at 2:42

Both these quotients have the following universal property: Let $X$ be a group (topological space). In the case of groups, you can associate a equivalence relation to every normal subgroup, which is $x\sim y$ if they belong to the same coset. Then $G/N$ is just $G/\sim$ as a set.

So now we just view the common problem of going mod an equivalence relation. The quotient structure on $X/\sim$ is one that makes the quotient map $p: X\to X/\sim$ have the universal property that given a map (continous or group homomorphism respectively) $g:X\to Y$ with $f(x)=f(y)$ for all $x\sim y$ then there exists a unique $\bar{g}:X/\sim \to Y$ such that $g=p\circ\bar{g}$.

You can try and workout that in the case of groups, this universal property makes sure that the quotient group structure on $G/N$ is the only one where $G\to G/\sim$ is a group homomorphism. But for groups we can't go mod arbitrary equivalence relations only those arising from normal subgroups.

On the other hand for topological spaces, there can be other topological space structures on $X/\sim$ which make $X\to X/\sim$ continuous, but only the quotient map has the universal property.

• To connect this to my answer, the functions $g$ satisfying your condition are exactly the functions which have the form $\varphi\circ f$ for some (necessarily unique) function $f$ on $X/{\sim}$. – Eric Wofsey Sep 2 '16 at 3:07
• @EricWofsey Yes! But of these are essentially the same universal property. You can use one to prove the other. – Arun Kumar Sep 2 '16 at 3:12
• Right, I didn't mean to imply otherwise--I just wanted to indicate for the benefit of readers how to relate them to each other. – Eric Wofsey Sep 2 '16 at 3:15
• Right, I got what you meant. I wanted to indicate that as well. My comment was supposed to be "both of these" – Arun Kumar Sep 2 '16 at 3:32