The following is from this paper.
A bordism $B:\coprod_{i\in I} S^1 \to \coprod_{j\in J} S^1$ is just a “generalized pair of pants” - a pair of pants with one waist hole for each $i\in I$ and one leg hole for each $j\in J$ Since both $I$ and $J$ are finite sets, as required by compactness....
I assume the author is using the general topology version of compactness, i.e., the existence of a finite subcover. However, I don't see how to apply compactness to get the finiteness of $i$ and $j$ conclusion. What is the open cover and what is the finite subcover in this case? The result seems to imply that the disjoint union of $S^1$ is the finite subcover but I can't seem to make this precise.
EDIT: This is actually fairly simple, not sure what I was thinking. Basically if $i,j$ aren't finite then compactness would be violated.
Take $X=\coprod_{i\in I} S_i$ (where each $S_i$ is homeomorphic to $S^1$) with $I$ infinite. An open cover of $X$ is $C=\{S_i\}_{i\in I}$. For any finite subset $D\subset C$ there is some $i\in I$ such that $S_i\not\in D$ and thus $D$ can't be a subcover. We have found a cover $C$ without finite subcover, and thus $X$ isn't compact.