Context: I have came across an example asking whether the quotient map $q: \mathbb{R} \to \mathbb{R}/(0,1)$ is open (we identify the interval $(0,1)$ with a point) and I managed to show that it is.
Now I am wondering if the following a bit more general statement is true:
Let $(A_i)_{i \in I}$ be a disjoint family of open sets in $X$. Let $q: X \to X/\{A_i ; i \in I\}$ be a quotient map where we identify each $A_i$ to its own point. Then $q$ is an open map.
My attempt: Let $U \subseteq X$ be an arbitrary open set in $X$. Let's define a subset of indices $J = \{j \in I | A_j \cap U \neq \emptyset\}$. We need to show that $q^{-1}(q(U))$ is open in $X$. But since $q^{-1}(q(U)) = U \cup \bigcup_{j \in J}A_j$ and all $A_j$ are open we have shown that $q$ is open.
Best Answer
That argument is correct, and generalises to the case of finitely many closed sets being identified to a point (more common than "pressed") and then the quotient map is closed.
Identifying open sets to a point, as in $\Bbb R{/}(0,1)$ is uncommon because we lose properties like $T_1$, so the resulting quotient is already non-metrisable etc. Closed sets (or even compact ones) is much more common in practice.