Structure of a profinite group as a condensed set with an action of an open subgroup Let $G$ be a profinite group and $H$ be an open subgroup. As a continuous $H$-topological space, we have $G=\coprod_{G/H} H$. Does this also hold as condensed sets, i.e. do we have an identification of $\underline{H}$-condensed sets $\underline{G}= \coprod_{G/H} \underline{H}$ ? If not, what can we say about the structure of $\underline{G}$ as an $\underline{H}$-condensed set ?
More generally, can we say something about the structure of a compactly generated topological group as a condensed set with an action of (the condensed group associated to) an open subgroup of finite index ?
 A: Yes, that is true. I'm not sure I'm explaining the step that's confusing you, but you are asking about a special case of the statement that the functor $S\mapsto\underline{S}$ from profinite sets to condensed sets commutes with finite coproducts. This is a formal consequence of the fact that the Grothendieck topology on profinite sets used in the definition of condensed sets allows such finite disjoint unions as covers.
Even more generally, if $G$ is any condensed group (for example $\underline{G}$ for a topological group $G$) and $H\subset G$ is a subgroup that is open (in the sense that for any profinite set $S$ with a map $S\to G$, the pullback $S\times_H G\subset S$ is an open subset of $S$ -- this is satisfied for $\underline{H}\subset \underline{G}$ for an open subgroup $H$ of a topological group $G$), then as a condensed $H$-set, one has $G\cong \bigsqcup_{G/H} gH$, where the index set $G/H$ is a discrete condensed set (i.e. just a set). Indeed, it is clear that $\bigsqcup_{G/H} gH$ injects into $G$ (on the level of presheaves, and thus after sheafification). But for any map $S\to G$ from a profinite set $S$, one has, set-theoretically, $S=\bigsqcup_{G/H} S\times_G gH$, and each term $S\times_G gH\subset S$ is open. Thus, this gives an open cover over which one gets a map to $\bigsqcup_{G/H} gH$, as desired.
