I'm self studying topology and came across the concepts of topology coherent with a family of subsets and weak topology generated by a family of spaces. Specifically:
Given a topological space $(X, \mathcal{T})$ and a family of $((S_{\alpha}, \mathcal{T}_{\alpha}))_{\alpha \in A}$ of subspaces of $X$ which cover $X$, then $\mathcal{T}$ is said to be coherent with this family iff $A \in \mathcal{T}$ iff $A \cap S_{\alpha} \in \mathcal{T}_{\alpha}$.
As I understand this is the topology coinduced in X by the inclusion maps. Now, the idea of weak topology generated by a family of spaces (also union topology?) seems to be the other way around, given a family of topological spaces $((X_{\alpha}, \mathcal{T}_{\alpha}))_{\alpha \in A}$ and letting $U = \bigcup_{\alpha \in A} X_{\alpha}$ we want to construct a topology $\mathcal{U}$ on $U$ such that $\mathcal{U}|_{\alpha} = \mathcal{T}_{\alpha}$, where $\mathcal{U}|_{\alpha}$ is the subspace topology induced on $X_{\alpha}$ by $\mathcal{U}$. This seems related to being coherent in the above sense but I don't know if it's equivalent. The topology $\mathcal{U}$ need not exist as discussed in this question: When do the coherent topology on $X$, induced by topologies $\{\mathcal{T}_i\}$ on a family of subsets $\{X_i\}$, agree with $\{\mathcal{T}_i\}$?
A necessary condition on the spaces $X_{\alpha}$ is:
- The subspace topologies on $X_{\alpha} \cap X_{\beta}$ induced by $X_{\alpha}$ and $X_{\beta}$ must coincide.
I understand this condition. However, there is (see Dugundji or Bourbaki) also
- $X_{\alpha} \cap X_{\beta}$ must be open or closed in both $X_{\alpha}$ and $X_{\beta}$.
I can't wrap my head around 2. I understand that together with 1 it is a sufficient condition but I can't think of how one could "naturally" arrive at it neither if it is also necessary. So my questions are:
Are there necessary and sufficient conditions on the family $X_{\alpha}$ such that there is a coherent topology on $U$? The books I've studied don't elaborate much on the weak topology generated by spaces.
Is there a "intuitive" argument on how to arrive at condition 2?
