The quotient of a quotient space/Quotient Topology

Question

Show that the quotient space of the quotient space of the set $X$ is a quotient space of $X$

I think that I should prove that the quotient set of $X$ is homeomorphic to the quotient set of the quotient set of $X$

How look likes the quotient of the quotient space?

What is the quotient map of the quotient of the quotient space?

Answer 1

If $X$ is a topological space and $R_1$ is an equivalence relation on $X$, one forms the quotient map $\pi_1:X\to X/R_1$, and equips $X/R_1$ with the quotient topology. That is to say, a subset of $X/R_1$ will be open, by definition, if its $\pi_1$-inverse image is open in $X$.

If $R_2$ is an equivalence relation on $X/R_1$, one repeats the process to obtain a second quotient map $\pi_2:X/R_1\to (X/R_1)/R_2$.

Thus the question becomes: is there a third equivalence relation $R_3$ on $X$ such that $(X/R_1)/R_2$ is homeomorphic to $X/R_3$?

It turns out that the answer is yes. What is the natural candidate to $R_3$? Well, one can say that $xR_3y$ if $\pi_1(x)R_2\pi_1(y)$ (which is equivalent to saying that $\pi_2(\pi_1(x))=\pi_2(\pi_1(y))$). Then it is easy to see that $R_3$ is indeed an equivalence relation. Hence we have $\pi_3:X\to X/R_3$.

And what is the homeomorphism? Follow your nose: $$X/R_3 \ni \pi_3(x)\mapsto \pi_2(\pi_1(x))\in (X/R_1)/R_2.$$

Answer 2

Taking the more abstract view:

$Y$ is a quotient space of $X$ iff there is a quotient map $q: X \to Y$, i.e. $q$ is onto and $$\forall O \subseteq Y: O \text { open in } Y \iff q^{-1}[O] \text { open in } X$$

Now, if $Z$ is a quotient of $Y$ we thus again have a quotient map $q': Y \to Z$ with $q'$ onto, and $$\forall O \subseteq Z: O \text { open in } Z \iff q'^{-1}[O] \text { open in } Y$$

Now just note that $q' \circ q: X \to Z$ is also onto, as the composition of two onto maps and $O \subseteq Z$ is open iff $q'^{-1}[O]$ is open in $Y$ iff $q^{-1}[q'^{-1}[O]]$ is open in $X$ and as $q^{-1}[q'^{-1}[O]] = (q' \circ q)^{-1}[O]$ we have shown that $q' \circ q$ is a quotient map. And so $Z$ is a quotient of $X$.

(some texts do not include the ontoness condition, but I find it more natural to include it).