Suppose $Q$ is a topological space and $Q/\sim$ the corresponding quotient space. Let $\pi: Q \rightarrow Q/\sim$ denote the projection map, mapping each $q \in Q$ to the corresponding equivalence class $[q] \in Q/\sim$.
I had initially thought that this map is open (or equivalently, that its inverse is continuous) by definition. A quick proof is suppose $U \subset Q/\sim$ is open. Then $\pi^{-1}(U)$ is open by definition, as the quotient topology is defined by taking all sets of the form $\pi(V)$ to be open where $V$ is in the topology of $Q$.
However based on a few questions I have read on here it seems that the inverse of the projection map is not necessarily continuous. Where does my proof fail? Also, am I mistaken to take a continuous inverse to be the same as an open map?
If I had to guess, perhaps my assumption that the quotient topology is generated by all the open sets via the projection map is incorrect? In other words, the topology on $Q/\sim$ is generated by all sets of the form $\pi(U)$, where $U$ is in the topology of $Q$. If so, what is the difference from this approach and that of only considering sets of the form $\pi^{-1}(U)$?
Best Answer
Your "proof" simply uses the definition of continuity for the map $\pi$. That is, you've simply shown why $\pi$ is continuous, not its inverse. However, don't confuse the notation for $\pi^{-1}(U)$, which stands for the pre-image of $U$ under $p$, for the inverse of a function. The notation $\pi^{-1}$ one sees in the definition of continuity always means the pre-image - it's not, in general, a function. We are not necessarily assuming existence of inverses in the definition of contunity. In fact, we can even say that in general, the projection map does not have an inverse. As an example, we can define an equivalence relation $\sim$ on $\mathbb{R}^2$ by letting $(x,y) \sim (x',y')$ iff $y = y'$. The map $p: \mathbb{R}^2 \to \mathbb{R}^2 / \sim$ defined by $p(x,y) = [y]$ will not have an inverse $g: \mathbb{R}^2 / \sim \to \mathbb{R}^2$. For instance, what would $g([2])$ be? It could be $(1,2)$. But it could just as likely be $(0,2)$ or $(-100, 2)$. The problem is that the projection map $p$ is not necessarily injective.
Now, if we assume the projection map is bijective (technically only injectivity is necessary since the projection map is surjective, by definition), we can ask whether or not your characterization would be correct. In fact, the answer is yes. I'll state this formally:
Proof: Let $g: Q / \sim \to Q$ be the inverse of the projection map $p: Q \to Q / \sim$, and assume $p$ is open. Now, let $U \subset Q$ be an open set. We must show $g^{-1}(U)$ is open. But thanks to the fact $p$ is bijective, we know that $g^{-1}(U) = p(U)$. As $p$ was an open map, by assumption, we conclude $p(U)$ is open. Thus, $g^{-1}(U)$ is open implying that the inverse function $g$ is continuous. To prove the converse, suppose that $g$ is continuous, and let $U \subset Q$ be open. By the bijectivity of $p$, we have $p(U) = g^{-1}(U)$, and as $g$ was continuous by assumption, $g^{-1}(U)$ is open $\implies p(U)$ is open. Thus, $p$ is an open map.