Projection of glueing identification is open map? Let $(X_\alpha)_{\alpha \in A}$ be a family of topological spaces and let there be given for every $\alpha,\beta \in A$ open subsets $X_{\alpha\beta} \subseteq X_\alpha$ and homeomorphism
$$ \psi_{\alpha\beta} \colon X_{\alpha\beta} \to X_{\beta\alpha}$$
such that $\psi_{\beta\alpha}^{-1} = \psi_{\alpha\beta}$ and $\psi_{\alpha\gamma} = \psi_{\beta\gamma} \circ \psi_{\alpha\beta}$ for all $\alpha,\beta,\gamma$ on appropriate domains.
Then we can glue the spaces together and obtain
$$ X := \bigl(\bigsqcup_{\alpha \in A} X_\alpha\bigr)\big/\sim$$
where $(x,\alpha) \sim (y,\beta)$ iff $x \in X_{\alpha\beta}$, $y \in X_{\beta\alpha}$ and $\psi_{\alpha\beta}(x) = y$. 
Question:  Is the natural projection $\pi \colon \bigsqcup_{\alpha \in A} X_\alpha \to X$ an open map?
(Then the sets $\pi(X_\alpha \times \{\alpha\})$ form an open covering of $X$, since the inclusion of $X_\alpha$ in the disjoint union is an open map. Also maybe we need to add $(x,\alpha) \sim (x,\alpha)$ manually, I'm not sure)
 A: Let $Y$ be the disjoint union and $p : Y \rightarrow X$ be the quotient map. Let $U$ be open in $Y$, we would like to show that $p(U)$ is open in $X$, that is, that its inverse image is open in $Y$. But $p^{-1}(p(U)) = p^{-1}\left( p\left( \cup_{\alpha\in A} X_{\alpha}\cap U\right) \right) = \cup_{\alpha\in A} p^{-1}\left( p\left(  X_{\alpha}\cap U\right) \right)$ and this equal to $\cup_{\alpha\in A} \cup_{\beta\in A} X_{\beta \alpha} \cap \psi_{\alpha \beta} \left( U\cap X_{\alpha \beta} \right)$. As $U$ is open in $Y$, so is $U\cap X_{\alpha}$ in $X_{\alpha}$, and (therefore) so is $U\cap X_{\alpha \beta}$ in $X_{\alpha}$, which implies that $X_{\beta \alpha} \cap \psi_{\alpha \beta} \left( U\cap X_{\alpha \beta} \right)$ is open in $X_{\beta\alpha}$ and in $X_{\beta}$, as $X_{\beta\alpha}$ is open in $X_{\beta}$. From this, one concludes that $\cup_{\alpha\in A} X_{\beta \alpha} \cap \psi_{\alpha \beta} \left( U\cap X_{\alpha \beta} \right)$ is open in $X_{\beta}$, and that $\cup_{\beta\in A} \cup_{\alpha\in A} X_{\beta \alpha} \cap \psi_{\alpha \beta} \left( U\cap X_{\alpha \beta} \right)$ is open in $Y$. This is but $\cup_{\alpha\in A} \cup_{\beta\in A} X_{\beta \alpha} \cap \psi_{\alpha \beta} \left( U\cap X_{\alpha \beta} \right)$, showing that it is open in $Y$, and that the quotient map $p$ is open.
