# Abstract Nonsense

## Quotient Topology (Structural Details)

In this post we shall discuss what I like to call the “structural details” of the quotient topology. By structural I mean that the construction of the quotient topology is unique in a lot of ways. This uniqueness is expressed in an almost categorical sense though. Namely, it all has to do with the following theorem:

Theorem: (Characteristic Property of Quotient Maps): Let $\xi:X\leadsto Y$. Then, $\zeta:Y\to Z$ is continuous if and only if $\zeta\circ\xi:X\to Z$ is continuous.

Proof: Clearly if $\zeta$ is continuous then so is $\zeta\circ\xi$ (since it is the composition of continuous maps).

Conversely, suppose that $\zeta\circ\xi$ is continuous and let $U\subseteq Z$ be open. Then, $(\zeta\circ\xi)^{-1}(U)=\xi^{-1}(\zeta^{-1}(U))$ is open in $X$. But, upon closer inspection this is exactly the condition that states that $\zeta^{-1}(U)$ is open in $Y$. The conclusion follows. $\blacksquare$

It is called the characteristic property of quotient maps because as we shall shortly see it really does characterize them. But first, we prove an important tool.

Theorem (Passing the Quotient): Let $\xi:X\leadsto Y$ and $\zeta:X\to Z$ be such that $\zeta(\xi^{-1}(\{y\}))$ is a singleton for each $y\in Y$ (i.e. it’s constant on the fibers). Then, there exists a unique continuous map $\overset{\sim}{\zeta}:Y\to Z$ such that $\zeta=\overset{\sim}{\zeta}\circ\xi$.

Proof: Since $\zeta(\xi^{-1}(\{y\}))$ is a singleton the map $\overset{\sim}{\zeta}:Y\to Z$ given by $\overset{\sim}{\zeta}:y\mapsto\zeta(\xi^{-1}(\{y\}))$ is well-defined (note, this really is an abuse of notation since the RHS is a set, but the intention is clear and the notation less messy). Also, with this definition $\zeta(x)=\zeta(\xi^{-1}(\xi(x)))=\overset{\sim}{\zeta}(\xi(x))$ and so $\zeta=\overset{\sim}{\zeta}\circ\xi$.

From there it is clear that $\overset{\sim}{\zeta}$ is continuous since it’s composition with the quotient map is continuous.

Lastly, the uniqueness follows from the fact that if $\theta:Y\to Z$ were another such map, then given any $\xi(x)\in Y$ we have that $\overset{\sim}{\zeta}(\xi(x))=\zeta(x)=\theta(\xi(x))$

We now move onto the meat of the post. Before we do we need to prove a nice little theorem which enables us to say that two spaces are identical.

Theorem: Let $(X,\mathfrak{J})$ and $(X,\mathfrak{S})$ be two topological spaces. Then, $\mathfrak{J}=\mathfrak{S}$ if and only if the identity map $\iota:(X,\mathfrak{J})\to (X,\mathfrak{S})$ is a homeomorphism.

Proof: If $\mathfrak{S}=\mathfrak{J}$ the conclusion is obvious.

Conversely, let $U\in\mathfrak{J}$ then $\iota(U)=U\in\mathfrak{S}$ since $\iota$ is open. Conversely, let $U\in\mathfrak{S}$ then $\iota^{-1}(U)=U\in\mathfrak{J}$ since $\iota$ is continuous.

The conclusion follows. $\blacksquare$.

We are now prepared to prove that a quotient map is the unique map which has the characteristic property described above. Then following it we shall use the above lemma to prove the uniqueness of a quotient space.

Theorem: Let $\xi:X\to Y$ be surjective. Then, $\xi:X\leadsto Y$ if and only if the characteristic property holds.

Proof: If $\xi:X\leadsto Y$ then this is the previous theorem.

Conversely, suppose that $\xi:X\to Y$ is a surjective map that has the characteristic property. Then, $\xi$ is continuous since $\iota\circ\xi:Y\to Y$ is continuous.

To show that $\xi$ is a quotient map it remains to show that $Y$ with it’s given topology, call it $Y_i$ is homeomorphic to $Y_q$ where $Y_q$ is obviously $Y$ with the quotient topology generated by $\xi$. Then, let $\xi_i:X\to Y_i$ and $\xi_q:X\to Y_q$ be the surjective map $\xi$ from $X$ to $Y_i,Y_q$ respectively.

We show that $\iota:Y_i\to Y_q$ is a homeomorphism, from where the conclusion will follow from the last theorem.

But, this is clear since $\xi_i=\iota\circ\xi_q$ and thus by the characteristic property $\iota$ is continuous. Conversely, $\iota^{-1}$ is continuous since $\xi_q=\iota^{-1}\circ\xi_i$.

The conclusion follows. $\blacksquare$

That is quite a profound statement, saying that really a quotient map is the manifestation of the unique kind of map such that the characteristic property holds.

We now prove an equally profound statement amounting to that in most cases if $Y,Y'$ are quotient spaces from the same map then $Y\approx Y'$. This will help us tremendously in our construction of explicit quotient spaces in our next post.

Theorem: Let $\xi_1:X\leadsto Y_1$ and $\xi_2:X\leadsto Y_2$ are quotient maps such that $\xi_1^{-1}(\xi_1(x))=\xi_2^{-1}(\xi_2(x))$ for every $x\in X$. Then, there exists a unique homeomorphism $\zeta:Y_1\to Y_2$ such that $\zeta\circ\xi_1=\xi_2$.

Proof: First let us examine $\xi_2:X\to Y_2$. Note, that $\xi_2(\xi_1^{-1}(\xi(\{x\})))=\xi_2(\xi_2^{-1}(\xi_2(\{x\})))=\xi_2(\{x\})$ and so $\xi_2$ is constant on the fibers of $\xi_1$ and so by passing the quotient we know there exists continuous map $\zeta_1:Y_1\to Y_2$ such that $\xi_2=\zeta_1\circ\xi_1$.

Similarly, using the exact same methodology we are furnished with a continuous map $\zeta_2:Y_2\to Y_1$ such that $\xi_1=\zeta_2\circ\xi_2$.

But, $\xi_1=\xi_1\circ\iota=\xi_1\circ(\zeta_1\circ\zeta_2)$ and thus by the uniqueness part in passing the quotient it follows that $\iota=_{Y_1}\zeta_1\circ\zeta_2$ and applying the same concept we see that $\iota_{Y_2}=\zeta_2\circ\zeta_1$. It follows that $\zeta_1$ is invertible with $\zeta_1^{-1}=\zeta_2$. But, since $\zeta_1,zeta_2$ are continuous this implies that $\zeta_1$ is a homeomorphism.

The last part is obvious. $\blacksquare$

This concludes our talk on structure.

Note: There are analgous structure theorems for the subspace, product, and disjoint union topologies. I have just not gotten the time to talk about them. The quotient topology ones come up most often and not surprisingly are the hardest to prove though.