This is a straightforward question, but not one that Hatcher clarifies:
Hatcher likes to write the disjoint union of topological spaces with infix $\coprod $. This works well enough, but there's a certain level of ambiguity interpreting his expression of form
\begin{align}
A \underset{i} \coprod B_i
\end{align}
which he enjoys using at very critical points in his book (such as when he constructs cellular complexes).
(Q.1): Have a finite index $i \in \left( I = \{ 1, 2, \cdots, n\}\right)$. Then does Hatcher imply
\begin{align}
A \underset{i} \coprod B_i = A \coprod B_1 \coprod B_2 \coprod \cdots \coprod B_n \quad \quad \quad \quad\; \; \; \quad \quad \quad \text{(Case 1)}
\end{align}
or does he strangely imply
\begin{align}
A \underset{i} \coprod B_i = \bigg(A \, B_1 \bigg) \coprod \bigg( A \,B_2 \bigg) \coprod \cdots \coprod \bigg(A \, B_n \bigg) \; \; \; \; \; \quad \; \text{(Case 2)}
\end{align}
I'd assume (Case 1) is true (I haven't seen him define concatenation) but I want to check since textbooks will switch from abusing notation (and not explaining when) to having definitions like compactness in which every word counts.
(Q.2) I prefer extending $+$ to topological spaces and using it to denote the coproduct. I'll also denote $\bigg( B_1 + B_2 + \cdots B_n = \sum \limits_i B_i \bigg)$. Then is it valid to rewrite Hatcher and have
\begin{align}
A \underset{i} \coprod B_i = A + \sum \limits_i B_i \quad \text{(Rewrite 1)}
\end{align}
If (Rewrite 1) holds, then there is a bit off oddness in Hatcher's abuse of notation, as you would have something analogous to stating $\bigg( A + \sum \limits_i B_i = A \sum \limits_i B_i \bigg) $. This could be reconciled by thinking of $ + B_i $ as being a string that you prolong $i$ times to form the expression.
Thanks in advance for your help.
Best Answer
Case 1 is the case. The notation is pretty much standard and I would not call it "abuse" by any measure. For reference, consider the following quote from page 5 of AT, right from the definition of cell complexes you mention: "Thus as a set, $X^n = X^{n - 1} \coprod_\alpha e_\alpha^n$ where each $e_\alpha^n$ is an open $n$-disk." Clearly the only sensible interpretation is as $X^n = X^{n - 1} \sqcup \coprod_\alpha e_\alpha^n$, as I am sure anybody familiar with CW-complexes will gladly confirm :)