Needing a clearer idea on “quotient topology”

So here's the definition of a quotient topology

Let $$(X,\tau)$$ be a toplogical space and define some equivalence relation $$\sim$$ on the set $$X$$. There exists a natural surjection denoted $$p:X \to X/\sim$$. Then the quotient topology on $$X/\sim$$ is

$$\tau_q=\{W \subseteq X/\sim : p^{-1}W \in \tau\}$$

Well, I am trying to see that $$\tau_q$$ is actually a topology. I am stuck on how to even think about the first axiom that "the entire set is in the topology."

Namely, I need to show that "$$X/\sim \in \tau_q$$" or equivalently,

$$p^{-1}(X/\sim) \in \tau$$

My problem is, $$X/\sim$$ is essentially a set of equivalence classes. So, if $$X \to X/\sim$$ I know exactly which equivalence class/element of $$X/\sim$$ it will be sent to. The reverse $$X/\sim \to X$$ is not so clear necessarily. That is one issue. Another is, the ambiguity of what topology is defined on $$X$$, namely, what are the elements of $$\tau$$? All I know is that they satisfy the requirements of a topology.

I consider the modulo $$2$$ on set $$\mathbb{Z}$$. Then $$\mathbb{Z}/\sim=\{[0],[1]\}$$. I can specify a discrete topology on $$\mathbb{Z}$$ for instance(but what if some other topology is defined on it?). So $$\tau$$ consists of all subsets(power set) of the integers. In this case, sine any subset imaginable on the integers is open so $$p^{-1}(\mathbb{Z}/\sim) \to \mathbb{Z}$$ is open.

I just don't know how to generalize this; for some other topology? Like say, how about the indiscrete topology, so $$\tau=\{\mathbb{Z},\phi\}$$. Then, $$p^{-1}$$ might send $$[0]$$ to any even number so, it could be $$0,\pm 2,\pm4...$$ and similarly to the odds for $$[1]$$.

Say it sends $$[0] \to 2$$ and $$[1] \to 1$$. So basically, $$\mathbb{Z}/\sim$$ is sent to the set $$\{1,2\} \in \mathbb{Z}$$. Is this in the topology? i.e. is this set open? Well no, because $$\{1,2\} \not\in \tau=\{\mathbb{Z},\phi\}$$.

I mean, it need not be sent to $$1,2$$ it could be anything else as long as they are even/odd respectively, but even so the "pair" of even and odd integers is not in $$\tau$$ the indiscrete topology on $$\mathbb{Z}$$.

So, $$p^{-1}\mathbb{Z}/\sim \not\in \tau$$ in this case and therefore the entire set $$\mathbb{Z}/\sim \not\in \tau_q$$, which violates the axiom a topology must satisfy.

where have I gone wrong in my thoughts? Please give me an idea of how to prove that $$\tau_q$$ is a topology!

$p^{-1}$ is not the inverse function, is the $\textbf{inverse image}$ of $\, p: X \to Y$, and is defined for every subset $W \subset Y$ of the codomain of $p$ as $$p^{-1}(W) := \{x \in X \mid p(x) \in W\}$$
Now, if $\, p: X \to X/\sim$, then $p$ is surjective, and so $$p^{-1}(X/\sim)= \{x\in X \mid p(x)\in X/\sim\}= X \in \tau \, \text{so } X/\sim \in \tau_q.$$ $$\text{What about } p^{-1}(\emptyset)?$$
The problem doesn't state anything particular of $\tau$, so you just know that it satisfies the axioms of a topology. A quick check on the properties of the inverse image should help you prove the union and intersection properties and hence showing that $\tau_q$ is in fact a topology for $X/\sim$.
For any function $f:A\to B$ between any sets, you always have $f^{-1}(B) = A$, because all elements of $A$ are sent into $B$.
Apply this to $p:X\to X/\sim$ : $p^{-1}(X/\sim) = X$.
• how about in the context of my example? Are you saying that $p^{-1}: \mathbb{Z}/\sim=\mathbb{Z}_2 \to \mathbb{Z}$ is a map $p^{-1}(\mathbb{Z}_2)=\mathbb{Z}$? How? – John Trail May 2 '16 at 17:16
• @JohnTrail Watch out: $p^{-1}$ is not a function. It is a short way of saying "all those $x$ such that $p(x)\ldots$" – Riccardo Orlando May 2 '16 at 22:44