# Abstract Nonsense

## Quotient Topology

In this post we discuss the very fruitful idea of a quotient topology (to be defined). The basic idea is that in our past ventures into the world of topology we have discussed many ways to form new topological spaces out of old ones. For example, if we have a non-empty class of non-empty topological spaces $\left\{X_j\right\}_{j\in\mathcal{J}}$ we can define a topology on $\displaystyle \prod_{j\in\mathcal{J}}X_j$ by defining an open subbase of the topology as $\displaystyle \bigcup_{j\in\mathcal{J}}S_j$ where $S_j=\left\{\pi_j^{-1}(U):U\text{ is open in }X_j\right\}$, in other words the product topology. Similarly, if $X$ is a topological space and $U\subseteq X$ we can topologize $U$ by using the subspace topology $\left\{U\cap O:O\text{ is open in }X\right\}$. Lastly, we have recently defined the topology on $\displaystyle \coprod_{j\in\mathcal{J}}X_j$ to be the set of all sets such that the inverse image of the sets under all the canonical injections $\varphi_j:x\mapsto (x,j)$ are open. Now, we discuss what is arguably the most important kind of topological construction… a quotient space. The idea was born of a topolgists wish to make rigorous the idea of “gluing” spaces or parts of spaces together.

For example, take the interval $[0,2\pi]$ and of connecting the two endpoints $0$ and $2\pi$. What you get is something that looks like $\mathbb{S}^1$. Another example would be of taking two unit circles and attaching them at a point to make something analogous to the figure eight space $\mathcal{E}$. Both these ideas have the common theme that we took two points and made them one. This can be thought of as, in the first example, defining a partition on $[0,2\pi]$ by $\mathcal{P}=\left\{\{x\}:x\in(0,2\pi)\right\}\cup\{0,2\pi\}$. For, now every point has remained the same except the endpoints of the interval have become “one point”.

Now that we have the motivation of a quotient topology let us begin:

Quotient Topology: Let $X$ be a topological space and $Y$ an untopologized set. Then, if $\xi:X\to Y$ is a surjective map we can define the quotient topology induced by $\xi$ on $Y$ by calling a subset $U$ of $Y$ open if and only if $\xi^{-1}(U)$ is open in $X$.

It is clear that if $Y$ has the quotient topology induced by $\xi$ that $\xi$ is a continuous surjective map. This does NOT mean that if $\varphi:X\to Y$ is a continuous surjective map that $Y$ has the quotient topology. For example, let $X=\{a,b\}$ and $X_1,X_2$ be $X$ topologized with the discrete and indiscrete topologies respectively. Then, the identity map $\iota:X_1\to X_2$ is a continuous surjective map but $Y$ does not have the quotient topology. To see this we must merely note that $\iota^{-1}(\{a\})=\{a\}$ which is open in $X_1$ but $\{a\}$ is not open in $X_2$. If $Y$ had the quotient topology then $U\subseteq Y$ would be open if and only if $\iota^{-1}(U)$ is open.

Remark: It makes sense from the formula $X-\xi^{-1}(U)=\xi^{-1}(Y-U)$ that a subset of a space with the quotient topology is closed if and only if its inverse image is closed.

It makes sense that in the definition $\xi$ is a “special” kind of map. We make this idea clear by the following definition

Quotient Map: Let $\xi:X\to Y$ be a continuous surjective map, then if $Y$ has the quotient topology induced by $\xi$ we call $\xi$ a quotient map.

Note: To save a lot of writing, this blog will utilize the convention that that $\rightsquigarrow$ indicates a quotient map. Thus, instead of saying “Let $\xi:X\to Y$ be a quotient map” we shall say “Let $\xi:X\leadsto Y$

We have yet to verify that the quotient topology actuall does induce a topology on $Y$ but this is a trivial task once one recalls that $\displaystyle \xi^{-1}\left(\bigcup_{j\in\mathcal{J}}U_j\right)=\bigcup_{j\in\mathcal{J}}\xi^{-1}(U_j)$.

So, we first wish to prove some equivalent definitions of quotient maps. But, before we do so we need a few definitions.

Saturated Set: Let $\varphi:X\to Y$ be any surjective map (here $X,Y$ are arbitrary untopologized sets). Then, we call $U\subseteq X$ saturated if $U=\varphi^{-1}(V)$ for some $V\subseteq Y$.

Fiber: Let $\varphi,X,Y$ be as above. Then, $\varphi^{-1}(\{y\})$ is called a fiber assuming $y\in Y$.

We now show how the two relate. But first, a slightly nice theorem.

Theorem: Let $\varphi, X,Y$ be as above and let $S\subseteq X$. Then, $S$ is saturated if and only if $S=\varphi^{-1}(\varphi(S))$.

Proof: Clearly if $S=\varphi^{-1}(\varphi(S))$ then it is saturated.

Conversely, if $S=\varphi^{-1}(V)$ for some $V\subseteq Y$ we see that $\varphi(S)=\varphi(\varphi^{-1}(\varphi(S)))=\varphi(S)$ (the last equality resulting from $\varphi$‘s surjectivity). The conclusion follows $\blacksquare$

Now we can prove how fibers and saturated sets are related.

Theorem: Let $\varphi,X,Y$ be as above and $S\subseteq X$. Then, $S$ is saturated if and only if $\displaystyle S=\bigcup_{y\in V}\varphi^{-1}(\{y\})$ for some $V\subseteq Y$.

Proof: If $\displaystyle S=\bigcup_{y\in V}\varphi^{-1}(\{y\})$ then $\displaystyle S=\bigcup_{y\in V}\varphi^{-1}(\{y\})=\varphi^{-1}\left(\bigcup_{y\in V}\{y\}\right)=\varphi^{-1}(V)$.

Conversely, if $S$ is saturated then $\displaystyle S=\varphi^{-1}(\varphi(S))=\varphi^{-1}\left(\bigcup_{s\in\varphi(S)}\{s\}\right)=\bigcup_{s\in\varphi(S)}\varphi^{-1}(\{s\})$. $\blacksquare$

Now, I know we’ve spent a fair amount of time talking about these saturated sets but the concept will be important. That said, we still have one more important theorem.

Theorem: Let $X,Y,\varphi$ be as above and $S\subseteq X$ saturated. Then, if $S\cap\varphi^{-1}(\{y\})\ne\varnothing$ then $\varphi^{-1}(\{y\})\subseteq S$.

Proof: Let $x\in\varphi^{-1}(\{y\})\cap S$ then $x\subseteq S$ and so $\varphi(x)=y\subseteq \varphi(S)$ and finally we may conclude that $\varphi^{-1}(\{y\})\subseteq\varphi^{-1}(\varphi(S))=S$. $\blacksquare$

Well, after all of this some of you may still believe that every subset of $X$ is saturated. The easiest way to intuitively reconcile yourself with this is to draw the obligatory diagram of ten dots mapping to two dots then pick one dot in the domain that maps to one dot in the codomain and another dot in the domain that maps to the other dot in the codomain. Convince yourself that this set is not saturated.

So, why all the fuss over these saturated sets? They provide a convenient alternative definition to a quotient map.

Theorem: Let $\xi:X\to Y$ be a continuous surjective map. Then, $\xi$ is a quotient map if and only if it takes saturated open or closed sets to saturated .

Proof: Let $\xi:X\leadsto Y$ and let $S\subseteq X$ be saturated open. Then, $\xi(S)$ is open because $\xi^{-1}(\xi(S))=S$ is open.

Conversely, suppose that $\xi:X\to Y$ takes saturated open sets to open sets. Now, it is clear that if $U\subseteq Y$ is open then $\xi^{-1}(U)$ is open (by continuity). Conversely, if $\xi^{-1}(U)$ is open then $U$ is open because $\xi^{-1}(U)$ is an open saturated set and thus by assumption so is $\xi(\xi^{-1}(U))=U$.

The case for closed saturated sets is identical. $\blacksquare$.

The above should clue you into the fact that in general a quotient map is not open (try to construct one for yourself!). That said, the converse of the satement is true.

Theorem: Let $\xi:X\to Y$ be a continuous surjective open or closed map. Then, $\xi$ is a quotient map.

Proof: Clearly if $U\subseteq Y$ is open then $\xi^{-1}(U)$ is open. Conversely, suppose that $\xi^{-1}(U)$ is open, then by assumption so is $\xi(\xi^{-1}(U))=U$.

The case for closed maps is analogous. $\blacksquare$.

Corollary: Let $\displaystyle X=\prod_{j\in\mathcal{J}}X_j$ be a product space and $\pi_j:X\to X_j$ the canonical projection, then $\pi_j$ is a quotient map.

With this and the following theorem (we have proved a stronger case of this before but it’s so important we’ll do it again!) we can characterize a lot of quotient maps.

Theorem: Let $X$ be a compact space and $Y$ a Hausdorff space, then if $\varphi:X\to Y$ is continuous and surjective it’s a quotient map.

Proof: Clearly if $U\subseteq X$ is open then so is $laetx \varphi^{-1}(U)$. Conversely, if $\varphi^{-1}(U)$ is open then $X-\varphi^{-1}(U)=\varphi^{-1}(X-U)$ is closed and thus compact. Thus $\varphi(\varphi^{-1}(X-U))=X-U$ is compact, but since $Y$ is Hausdorff this implies that $X-U$ is closed. Therefore, $U$ is open. $\blacksquare$.

As a quick side remark, if one combines this with the following theorem one gets a slick proof that every continuous bijection from a compact space onto a Hausdorff space is a homeomorphism.

Theorem: Let $\xi:X\leadsto Y$ be injective. Then, $\xi:X\leadsto Y$ is a homeomorphism.

Proof: Since $\xi$ is bijective it remains to show that it’s open. But, this follows since if $U\subseteq X$ is open then so $\xi(U)$ since $U=\xi^{-1}(\xi(U))$. $\blacksquare$

This says that is a sense a quotient map is the closest one can get to a homeomorphism without it actually being one.

Another reason that these saturated sets are important is that they are the subsets of $X$ for which the restriction of $\xi$ to is still a quotient map.

Theorem: Let $\xi:X\leadsto Y$ and $S\subseteq Y$ be an open or closed saturated set. Then, $\xi\mid_S:S\to\xi(S)$ is a quotient map.

Proof: We first need a small lemma.

Lemma: Let $V\subseteq \xi(S)$, then $\xi\mid_S^{-1}(V)=\xi^{-1}(V)$.

Proof: This follows since $\xi\mid_S^{-1}(V)=\xi^{-1}(V)\cap S$. But, since $V\subseteq \xi(S)$ this implies that for each $v\in V$ we have that $\xi^{-1}(\{v\})\cap S\ne\varnothing$. But, by previous theorem this implies that $\xi^{-1}(\{v\})\subseteq S$. Thus, $\displaystyle \xi^{-1}(V)=\bigcup_{v\in V}\xi^{-1}(\{v\})\subseteq S$. And by previous observation the conclusion follows. $\blacksquare$

So, now clearly $\xi\mid_S:S\to\xi(S)$ is continuous and surjective and so we must only prove that if $\xi\mid_S^{-1}(V)$ is open in $S$ then $V$ is open in $Y$. But, clearly since $\xi\mid_S^{-1}(V)$ is open in $S$ and $S$ open in $X$ this implies that $\xi\mid_S^{-1}(V)=\xi^{-1}(V)$ is open in $X$. Thus, since $\xi$ is a quotient map this implies that $V$ is open in $Y$ and thus $V\cap\xi(S)=V$ is open in $\xi(S)$. $\blacksquare$

It is also not hard to fathom that the composition of quotient maps is a quotient map.

Theorem: Let $\xi:X\leadsto Y$ and $\zeta:Y\leadsto Z$ then $\zeta\circ\xi:X\leadsto Z$.

Proof: Clearly $\zeta\circ\xi$ is continuous and surjective, thus it remains to show that $(\zeta\circ\xi)^{-1}(V)$ is open in $X$ implies that $V$ is open in $Z$. But, this clear since $(\zeta\circ\xi)^{-1}(V)=\xi^{-1}(\zeta^{-1}(V))$ and so it follows from definition that $\xi^{-1}(\zeta^{-1}(V))$ is open in $X$ implies that $\zeta^{-1}(V)$ is open in $Y$ which in turn implies that $V$ is open in $Z$. $\blacksquare$

We are winding down here for this first post on the quotient topology. We prove some assorted theorems. But, first we made one important observation.

Remark: It does not warrant the title proof, but any reader should make due note that since a quotient space is the continuous image of something any property held by the domain of a quotient map which is invariant under continuous maps is held by the image.

Now for some odds-and-ends proofs.

For the first one I’ll ask you to remember that second countability is invariant under open surjective maps. But, quotient maps need be open, right? Thus, it is not always true that a quotient space of a second countable space is second countable. But! There is a stronger condition that one can pose that does guarantee it!

Theorem: Let $\xi:X\leadsto Y$, $X$ second countable, and $Y$ locally Euclidean of dimension $n$. Then, $Y$ is second countable.

Proof: It was proven earlier that a locally Euclidean space has an open base of Euclidean balls. So, let $\Omega$ be just that open base in $Y$. Then, clearly $\left\{\xi^{-1}(U):U\in\Omega\right\}$ is an open cover for $X$. But, by assumption (since every second countable space is Lindelof) this has a finite subcover. So, let $\Sigma$ be the set of all those sets in $\Omega$ whose inverse image lie in the countable subcover. Then, $\Sigma$ is an open cover for $X$ since

$\displaystyle Y=\varphi(X)=\varphi\left(\bigcup_{U\in\Sigma}\varphi^{-1}(U)\right)=\bigcup_{U\in\Sigma}\varphi(\varphi^{-1}(U))=\bigcup_{U\in\sigma}U$

But, since each $U\subseteq\sigma$ is a homeomorphic to an open subspace of $\mathbb{R}^n$ it has a countable open base $B_U$. Clearly though since $U$ is open in $X$ each element of $B_U$ is also open in $X$. So, let $\displaystyle \Lambda=\bigcup_{U\in\Sigma}B_U$. We clearly have $\Lambda$ is countable since it is the countable union of countable sets. Thus, it remains to show that $\Lambda$ is an open base for $Y$. So, let $x\in Y$ be arbitrary and $N$ any neighborhood of it. By assumption $x\in U_0$ for some $U_0\in\Sigma$. Thus, $x\in U_0\cap N$ which is an open subspace of $U_0$. Thus, there exists some $B\in B_{U_0}\subseteq\Lambda$ such that $x\in B\subseteq U_0\cap N\subseteq N$. Since $x$ and $N$ were arbitrary the conclusion follows. $\blacksquare$

Theorem: Let $\xi:X\to Y$ be continuous with a continuous right inverse $\overline{\xi}:Y\to X$ such that $\xi\circ\overline{\xi}=\iota_Y$. Then, $\xi:X\leadsto Y$.

Proof: Surjectivity is a simple result from set theory. The fact that it is a quotient map follows since if $\xi^{-1}(V)$ is open then $\overline{\xi}^{-1}(\xi^{-1}(V))$ is open in $Y$ (since $\overline{\xi}$  is continuous). But $\overline{\xi}^{-1}(\xi^{-1}(V))=\left(\xi\circ\overline{\xi}\right)^{-1}(V)=\iota^{-1}(V)=V$. The conclusion follows. $\blacksquare$.

April 6, 2010 -

1. […] one last example of a topological manifold. We define the real projective space  to be the quotient space  where if for some . In other words, we can think about as being the space of one-dimensional […]

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2. f : X —> Y, C a saturated closed subset of X, f surjective and continuous. Therefore X – C is open. Is it saturated?

Nice site !

Comment by luysii | September 11, 2012 | Reply

• Thank you! No, this is not true from a purely set-theoretic point of view. Namely, merely find two sets $X,Y$ and a function $f:X\to Y$ such that there exists $C=f^{-1}(f(C))\subseteq X$ such that $f^{-1}(f(X-C))\ne X-C$ (I assure you, there exists one). Then, just give everything in sight the discreet topology and you’ll have found your counterexample.

Best of luck 🙂

Alex

Comment by Alex Youcis | September 12, 2012 | Reply

3. […] of this, as well as a lot of other cool math, in [6]). Consider then the quotient group with the quotient topology inherited by the canonical projection –note that is connected, being the quotient of a […]

Pingback by Riemann Surfaces (Pt. II) « Abstract Nonsense | October 2, 2012 | Reply

4. […] we can identify with . The advantage of this is that we now have a projection map , which using the quotient topology given by this map, allows us to topologize […]

Pingback by Riemann Surfaces (Pt. III) « Abstract Nonsense | October 2, 2012 | Reply