Why don't we simply say that the open sets in the quotient topology are projection of open sets of initial topology?

Question

When we define the quotient topology, we say that an open set in it, are those sets which have pre image as an open set. But, why not just define it to be the image of open set of the initial set?

Intuitively speaking, I guess it has something to do with wanting open sets to be quotiented with each other. I came to this idea by considering a toy example which I describe below.

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Consider the set $\{1,2,3\}$ with the topology $\tau = \{ \phi,\{1,2,3\},\{1\} \}$. Now, suppose we define a relation $~$ which has only a single element in the relation $1~2$, we turn this into an equivalence relation and quotient the set with it. We get the quotient set $\{[1], [3]\}$ and the corresponding quotient topology $\tau' = { \phi ,\{[1], [3]\}}$. We don't have $\{1\}$ any more because its pre image is $\{1,2\}$.

However, we can also check that, were it so that we had ${2}$ and ${1,2}$ in our initial topology, then it would be that $[1]$ by itself would also be an open set in the quotient.

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Is this the correct idea? If yes, how do I make this more precise, if no, what then is?

Answer 1

You want the projection onto the quotient to be continuous and to be universal relative to the quotient.

That is, we want the quotient $X/\sim$ and the map $p\colon X\to X/\sim$ to satisfy the universal property:

1. $p$ is continuous; and
2. If $Y$ is any topological space and $f\colon X\to Y$ is a continuous function such that for all $a,b\in X$, if $a\sim b$ then $f(a)=f(b)$, then there exists a unique continuous function $\overline{f}\colon X/\sim \to Y$ such that $f = \overline{f}\circ p$.

In order for this to occur, we need $X/\sim$ to have the finest topology that makes $p$ continuous. This because if $\tau$ and $\tau'$ are topologies on $X/\sim$ with $\tau\subsetneq\tau'$, then the identity map $\mathrm{id}\colon (X/\sim,\tau)\to(X/\sim,\tau')$ is not continuous. So we must ensure that there is no finer topology on $X/\sim$ that can make $p$ continuous.

That forces our hand: we must have that the open sets in $X/\sim$ must be precisely all subsets $A$ of $X/\sim$ such that $p^{-1}(A)$ is open.

By contrast, asking that the direct image of open sets be open does not ensure either that $p$ is continuous, nor the quotient have the universal property.

Answer 2

This is simply not true, that would imply that quotient maps are open, which is not the case. See:

Example of quotient mapping that is not open

What you can do is the following.

Definition: Let $X$ be a space, $E$ an equivalence relation on $X$. A set $A\subseteq X$ is invariant if whenever $A$ intersects some $E$-equivalence class, it contains the entire equivalence class (equivalently, if $\pi$ is the quotient map, $\pi^{-1}(\pi(A))=A$).

We then have that the quotient topology on $X/E$ is the topology in which the open sets are exactly the projections of the invariant open sets under the quotient map.

Answer 3

I wonder...

• Who has never wandered about that? :-)

Notice that any topology that makes the projection open contains the quotient topology. But the projection need not be open for it to be continuous.

This shows that if the projection is open in the quotient topology, then indeed, in this case, the quotient topology is the weakest topology that makes the projection open.

Once the map is open, adding more sets to the co-domain topology cannot turn it into a non-open map. The strongest the topology, the easiest for it to be open. Continuity, on the other hand, works on the opposite way. The more open sets you have on the co-domain the less likely the map will be continuous. So, when the quotient topology is such that the projection is open, you are in this very special situation where weaker topologies have continuous projection, stronger topologies have open projections and the only topology where the projection is both is (in this case) the quotient topology.

I guess this is what confuses us (me, at least).

For the projection to be open in the quotient topology, you need the saturation of any open set $A$, $$\pi^{-1}(\pi(A)),$$ to be open. So, if you pick up an open set $A$ and a set $B$ such that $A \cup B$ is not open, and collapse them both into one point $p$, the set $\{p\}$ will be the image of $A$, but it will not be open, because $\pi^{-1}(p)$ is the saturation $A \cup B$, which is not open.