Kerodon

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

is a pullback square (Remark 7.7.4.3). Note that (7.87) admits an essentially unique extension to a commutative diagram

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. $\square$

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

For each $i \neq j$, we can choose a commutative diagram

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

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). $\square$

Proof. We will give the proof for $\operatorname{\mathcal{QC}}$; the analogous statement for the $\infty $-category $\operatorname{\mathcal{S}}$ is similar (but easier). By virtue of Variant 7.7.4.8, the $\infty $-category $\operatorname{\mathcal{QC}}$ has disjoint coproducts. It will therefore suffice to show that small coproducts in $\operatorname{\mathcal{QC}}$ are universal. Let $\{ \operatorname{\mathcal{C}}_ i \} _{i \in I}$ be a small collection of $\infty $-categories and let $\operatorname{\mathcal{C}}= \coprod _{i \in I} \operatorname{\mathcal{C}}_ i$ denote their coproduct, formed in the category of simplicial sets. For each $i \in I$, let $f_ i: \operatorname{\mathcal{C}}_ i \hookrightarrow \operatorname{\mathcal{C}}$ be the inclusion map. The collection $\{ f_ i \} _{i \in I}$ determines a morphism $f: \{ \operatorname{\mathcal{C}}_ i \} _{i \in I} \rightarrow \underline{\operatorname{\mathcal{C}}}$ in the $\infty $-category $\operatorname{Fun}(I, \operatorname{\mathcal{QC}})$, which exhibits $\operatorname{\mathcal{C}}$ as a coproduct of $\{ \operatorname{\mathcal{C}}_ i \} _{i \in I}$ in the $\infty $-category $\operatorname{\mathcal{QC}}$ (Example 7.6.1.17).

Suppose we are given a functor of $\infty $-categories $\operatorname{\mathcal{C}}' \rightarrow \operatorname{\mathcal{C}}$. For each $i \in I$, let $\operatorname{\mathcal{C}}'_{i}$ denote the inverse image of $\operatorname{\mathcal{C}}_{i}$ in $\operatorname{\mathcal{C}}'$, so that we have a pullback diagram

in the category of simplicial sets. Since $f_{i}$ is an isofibration, this diagram is also a categorical pullback square (Corollary 4.5.2.27), and therefore determines a pullback diagram in the $\infty $-category $\operatorname{\mathcal{QC}}$ (Example 7.6.3.4). Allowing $i$ to vary, we obtain a levelwise pullback diagram

in the $\infty $-category $\operatorname{Fun}(I, \operatorname{\mathcal{QC}})$. To complete the proof, it will suffice to show thatt $f'$ exhibits $\operatorname{\mathcal{C}}'$ as a coproduct of $\{ \operatorname{\mathcal{C}}'_ i \} _{i \in I}$ in the $\infty $-category $\operatorname{\mathcal{QC}}$, which follows from Example 7.6.1.17. $\square$