Proposition 7.7.4.10. Let $I$ be a set and let $\operatorname{\mathcal{C}}$ be an $\infty $-category which admits pullbacks and $I$-indexed coproducts. Assume that $I$-indexed coproducts in $\operatorname{\mathcal{C}}$ are universal and that $\operatorname{\mathcal{C}}$ has a universal initial object $\emptyset $. Then the coproduct of a collection of objects $\{ X_ i \} _{i \in I}$ is strongly universal (in the sense of Definition 7.7.2.15) if and only if it disjoint (Definition 7.7.4.1).
$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$
Proof. Choose a collection of morphisms $f_ i: X_ i \rightarrow X$ which exhibit $X$ as a coproduct of $\{ X_ i \} _{i \in I}$. Using Corollary 7.7.2.19, we see that the coproduct is strongly universal if and only if the following condition is satisfied:
- $(\ast )$
Suppose we are given a morphism $u: Y \rightarrow X$ and a collection of diagrams $\sigma _ i$:
\[ \xymatrix { Y_ i \ar [r]^{g_ i} \ar [d] & Y \ar [d]^{u} \\ X_ i \ar [r]^{ f_ i } & X } \]in the $\infty $-category $\operatorname{\mathcal{C}}$. If the morphisms $g_ i$ exhibit $Y$ as a coproduct of $\{ Y_ i \} _{i \in I}$, then each $\sigma _ i$ is a pullback square.
Assume first that condition $(\ast )$ is satisfied. Fix an element $j \in I$, and let $\sigma _ j$ denote the commutative diagram
\[ \xymatrix { X_ j \ar [r]^{\operatorname{id}} \ar [d]^{\operatorname{id}} & X_ j \ar [d]^{ f_ j } \\ X_ j \ar [r]^{ f_ j } & X. } \]
For each $i \neq j$, we can choose a commutative diagram
\[ \xymatrix { \emptyset \ar [r] \ar [d] & X_ j \ar [d]^{ f_ j } \\ X_ i \ar [r]^{ f_ i } & X. } \]
Applying $(\ast )$ to the collection $\{ \sigma _ i \} _{i \in I}$, we conclude that each $\sigma _ i$ is a pullback square. In the case $i = j$, we conclude that $f_ j$ is a monomorphism (Remark 9.3.4.18), and for $i \neq j$ we conclude that the fiber product $X_ i \times _{X} X_ j$ is an initial object of $\operatorname{\mathcal{C}}$. Allowing $j$ to vary, we conclude that the morphisms $\{ f_ i: X_ i \rightarrow X \} $ exhibit $X$ as a disjoint coproduct of $\{ X_ i \} _{i \in I}$.
Now assume that the morphisms $f_{i}$ exhibit $X$ as a disjoint coproduct of $\{ X_ i \} _{i \in I}$, and suppose we are given a collection of commutative diagrams $\{ \sigma _ i \} _{i \in I}$ as in $(\ast )$. For each index $i \in I$, the diagram $\sigma _ i$ determines a comparison map $u: Y_ i \rightarrow X_ i \times _{X} Y$, and we wish to show that $u$ is an isomorphism. For each $j$, let $g'_{j}: X_ i \times _{X} Y_ j \rightarrow X_ i \times _{X} Y$ be the morphism obtained from $g_ j$ by pullback along $f_{i}$. Then $u$ factors as a composition
\[ Y_ i \xrightarrow {u'} X_ i \times _{X} Y_{i} \xrightarrow { g'_{i} } X_ i \times _{X} Y, \]
and our assumption that $f_ i$ is a monomorphism guarantees that $u'$ is an isomorphism (see Variant 9.3.4.20). Consequently, to show that $u$ is an isomorphism, it will suffice to show that $g'_{i}$ is an isomorphism. Since $I$-indexed coproducts are universal, the morphisms $\{ g'_{j} \} _{j \in I}$ exhibit $X_ i \times _{X} Y$ as a coproduct of the collection $\{ X_ i \times _{X} Y_ j \} _{j \in I}$. It will therefore suffice to show that, for $j \neq i$, the fiber product $X_ i \times _{X} Y_ j$ is an initial object of $\operatorname{\mathcal{C}}$. Since $X_ i \times _{X} Y_ j$ admits a morphism to the initial object $X_ i \times _{X} X_ j$, this follows from our assumption that the initial object of $\operatorname{\mathcal{C}}$ is universal (Example 7.7.1.17). $\square$