# Defining the quotient topology from the quotient map

I am a bit confused about the quotient topology.

If $p:X \rightarrow A$ is the quotient map from some topological space $X$ to a set $A$, then any subset $U \subset A$ is defined to be open in $A$, if $p^{-1}(U)$ is open in $X$.

To show this is a topology one needs to show $p^{-1}(\emptyset)$ and $p^{-1}(A)$ are both open in $X$, but I am not sure how we know this is the case.

I do not understand why it can never be the case that $p$ could map an element of $X$ to the empty set.

I would greatly appreciate any clarification you could offer.

## 2 Answers

By definition, $p$ is a well defined function. This means that for any $x\in X$, there must exist a unique $a\in A$ such that $p(x)=a$. So p could not map an element of X to an empty set.

$p^{-1}[\emptyset]$ by definition equals $\{x \in X: p(x) \in \emptyset\}$ and no such $x$ can exist as the empty set has no elements at all. So $p^{-1}[\emptyset] = \emptyset$

$p^{-1}[A] =\{x \in X: p(x) \in A\}$ and so this holds for all $x \in X$: $p(x)$ always is some well-defined value in $A$. So $p^{-1}[A] = X$ and $\emptyset$ and $X$ are both open in $X$.