Definition and construction of a quotient space using a toy example

Question

To make headway into Agebraic Topology I need to precisely understand the definition and construction of the quotient space. Definitions in my textbooks and online have felt handwavy to me, and I don't feel confident in understanding not only the definition of the quotient space, but what I am allowed to use for my equivalence relation and partition in the quotient set. My question is roughly split into three sub-parts, using a toy example to try and illustrate which parts of the construction I feel I need to know more about:

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Let's construct a quotient set:

Take a set given as $A = \{ x, y, z \} $. Define a simple equivalence relation $\sim$ as being: $$\star \sim \star’ \iff (\star = \star’ = y) \lor (\star \neq y \neq \star’) $$ Then (using a representative of each partition, each element maps as such): $$ y \mapsto [y] = \{y \}$$ $$x , z, \mapsto[x] = \{x, z \} $$

Then the quotient set $A \over \sim$ equals $\{ [x], [y] \}$

a) Is this a valid quotient set?

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Let's apply the final topology on the quotient set:

Let's look beyond sets and have a topology applied on both $A$ and $A \over \sim$. We'll represent our two topological spaces and the map $f$ between them as $(A, \Phi) \xrightarrow{\quad f \quad} ({A \over \sim}, \Psi)$. For the map to be continuous the following condition must hold:

$$ * \in \Psi \Longrightarrow f^{-1}(*) \in \Phi$$

b) Is the inverse defined as $f^{-1}[\star] = \star$? If so, does the choice of representative for each equivalence class matter? Will I get different results if I change $[x]$ to $[z]$?

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Giving this a quick look, it seemed that there was no topology $\Psi$ that could work if I decided to make $\Phi$ the trivial topology. However, if I made $\Phi$ and $\Psi$ both the discrete topology, then I obtained:$$ \Phi = \{ \{ \varnothing \} , \{ x, y, z\} , \{ x \} , \{ y \} , \{ z \} , \{ x, y \} , \{x, z\} , \{y, z \} \}$$ $$ \Psi = \{ \{ \varnothing \}, \{ [x], [y] \}, \{ [x]\}, \{ [y] \} \}$$

In this case, the inverse mapping seems to be continuous, and the discrete topology is obviously the finest.

c) Do we generally start with the topology $\Phi$ on $A$ and then "determine" from there the final topology $\Psi$ on $A \over \sim$? Are there some topologies on $A$ (such as the trivial topology) that can make a final topology (ergo a quotient space) impossible? Is my toy example a correct example?

Cheers, and thanks for the help.

Answer 1

I'll answer (c) first. Yes, we start with the topology $\Phi$ on $A$ and then uniquely determine from there the final topology $\Psi$ on $A \over \sim$. The quotient topology always exists and is unique (up to a homeomorphism). For example, if $A$ has the trivial topology then $A \over \sim$ also has the trivial topology.

As we all know, equivalence relation on, surjective map out of, and partition of a set are 3 equivalent descriptions from which we can define a quotient set. Indeed, if the set is a topological space, then from any one of them we could construct a unique (up to a homeomorphism) quotient space with the quotient topology.

(a) In your toy example, $\star = y$ is not an equivalence relation: an equivalence relation is binary. However, I understand what you mean, you want to partition the set $A$ into two equivalence classes: $[y]$ is the singleton $\{y\}$, $[x]$ contains all the remaining elements. Then the set $\frac{A}{\sim} =\{ [x], [y] \}$ is a valid quotient set, but we have yet to define its topology. The equivalence relationship on $A$ can be defined as $$a\equiv b\iff a=b=y\quad or \quad a\ne y,b\ne y$$

The surjective map $p:A\to \frac{A}{\sim}$ is defined as $$p(y)=[y],p(x)=[x],p(z)=[x]$$ (b) the inverse is defined as $$p ^{-1}([y])=\{y\},p ^{-1}([x])=\{x,z\}$$ $p^{-1}$ is not a function, it's just the inverse of a function. It sends an equivalence class in $\frac{A}{\sim}$ to all elements in $A$ belonging to that class. It's true that $p^{-1}$ sends an open set in $\frac{A}{\sim}$ to an open set in $A$, but this isn't all. It sends any subset of $\frac{A}{\sim}$ to all of their elements in $A$. The choice of representative for each equivalence class does not matter.

Coming back to (c), how do we construct that unique quotient topology? Here's an excerpt from Munkres:

Definition. If $X$ is a space and $A$ is a set and if $p : X \to A$ is a surjective map, then there exists exactly one topology $\tau$ on $A$ relative to which $p$ is a quotient map; it is called the quotient topology induced by $p$. The topology $\tau $ is defined by letting it consist of those subsets $U$ of $A$ such that $p ^{-1}(U)$ is open in $X$.

It is easy to check that $\tau $ is a topology. The sets $\emptyset $ and $A$ are open because $p ^{-1}(\emptyset ) = \emptyset $ and $p ^{-1}(A) = X$. The other two conditions for a topology are also easy to check, which I leave to you as an exercise.