Proposition 7.7.4.9. Let $\operatorname{\mathcal{C}}$ be an $\infty $-category which has disjoint coproducts. Suppose that $\operatorname{\mathcal{C}}$ also admits $I$-indexed coproducts, for some set $I$. Then every collection of objects $\{ X_ i \} _{i \in I}$ of $\operatorname{\mathcal{C}}$ admits a disjoint coproduct.
$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$
Proof. Since $\operatorname{\mathcal{C}}$ admits $I$-indexed coproducts, we can choose an object $X \in \operatorname{\mathcal{C}}$ and a collection of morphisms $\{ f_ i: X_ i \rightarrow X \} _{i \in I}$ which exhibits $X$ as a coproduct of $\{ X_ i \} _{i \in I}$. We first claim that, for each $i \in I$, the morphism $f_ i$ is a monomorphism. Note that the collection $\{ X_ j \} _{j \neq i}$ also has a coproduct $Y \in \operatorname{\mathcal{C}}$ (since it can be written as $I$-indexed coproduct, where we take one summand to be the initial object $\emptyset \in \operatorname{\mathcal{C}}$). The morphisms $\{ f_ j \} _{j \neq i}$ then determine a map $g: Y \rightarrow X$, and the morphisms $f_ i$ and $g$ exhibit $X$ as a coproduct of $X_ i$ with $Y$. Our assumption that $\operatorname{\mathcal{C}}$ has disjoint coproducts then guarantees $f_ i$ and $g$ are monomorphisms.
To complete the proof, it will suffice to show that if $i$ and $j$ are distinct elements of $I$, then every commutative diagram
7.87
\begin{equation} \begin{gathered}\label{equation:infinitary-disjoint-coproducts} \xymatrix { \emptyset \ar [r] \ar [d] & X_ i \ar [d]^{f_ i} \\ X_ j \ar [r]^{ f_ j } & X } \end{gathered} \end{equation}
is a pullback square (Remark 7.7.4.3). Note that (7.87) admits an essentially unique extension to a commutative diagram
\[ \xymatrix { \emptyset \ar [rr] \ar [dd] & & X_ i \ar [dd]^{f_ i} \ar [dl] \\ & X_ i \amalg X_ j \ar [dr]^{h} & \\ X_ j \ar [rr]^{ f_ j } \ar [ur] & & X, } \]
where the upper left region is a pushout square. Our assumption that coproducts in $\operatorname{\mathcal{C}}$ are disjoint guarantees that the upper left region is also a pullback square. Arguing as above, we see that $h$ exhibits $X_ i \amalg X_ j$ as a summand of $X$, and is therefore a monomorphism. The desired result now follows from Variant 9.3.4.20. $\square$