$X$ coherent with family $\mathcal{B}$ of subspaces.Then $\bigsqcup_{B\in\mathcal{B}} B\to X$ induced by inclusion of each $B\to X$ is a quotient map 
Proposition $5.2 \ (b)$: Suppose $X$ s a topological space whose topology is coherent with a family $\mathcal{B}$ of subspaces.
Then the map $\bigsqcup_{B \in \mathcal{B}} B \to X$ induced by the inclusion of each set $B \to X$ is a quotient map

(This is proposition 5.2 (b) in Introduction to Topological Manifolds by Lee)
What does this mean rigorously? I'm having a hard time trying to unwind exactly what the author is trying to say.
What I think the author is trying to say is that the map $f : \bigsqcup_{\alpha \in \mathcal{A}} B_{\alpha} \to X$ defined by $f(x, \alpha) = x$ where $\alpha \in A$ for some indexing set $A$ which is bijective with $\mathcal{B}$ (the reason for this is that $\bigsqcup_{B \in \mathcal{B}} B$ is then isomorphic to $\bigsqcup_{\alpha \in A} B_{\alpha}$ in the category of sets) is a quotient map.
Am I correct? Because $\bigsqcup_{B \in \mathcal{B}} B \to X$ isn't even defined because $\mathcal{B}$ is a family of subspaces (and hence sets) but it is not an indexed family of subspaces (and therefore not an indexed family of sets) so we can't form the abstract (set-theoretic) disjoint union.
 A: I think you are confusing yourself over a small notational issue.

Then the map $f \colon \bigsqcup_{B \in \mathcal B} B \to X$ induced by the inclusion of each set $B \to X$ is a quotient map.

Define the disjoint union to be
$$\bigsqcup_{B \in \mathcal B} B = \{ x_B \mid x \in B \text{ for some } B \in \mathcal B \}.$$
The points in the disjoint union thus "remember" what set in $\mathcal B$ they came from. A single point $x \in X$ now has a copy of itself in the disjoint union for each set $B \in \mathcal B$ in which it appears. The map $f$ identifies all these points again, taking $x_B$ to $x$. (That is, if $x \in B \cap B'$ for $B, B' \subset X$, then $f (x_B) = f (x_{B'}) = x$.)
A: This means what you think it means.  There is not actually any abuse of notation here, though.  It is completely standard to write something like $$\bigsqcup_{i\in I}f(i)$$ where $f$ is some function that takes an element $i\in I$ and outputs a set $f(i)$.  The indexed family you are taking the disjoint union of here is $(f(i))_{i\in I}$.  In the case of $$\bigsqcup_{B \in \mathcal{B}} B,$$ $I=\mathcal{B}$ and $f$ is the identity function, so $f(B)$ is just $B$.  In other words, this family is indexed, namely by the identity map $\mathcal{B}\to\mathcal{B}$.
