Proposition 4.7.3.7. Let $\kappa $ and $\lambda $ be cardinals, where $\lambda $ is infinite. The following conditions are equivalent:
The cardinal $\kappa $ is strictly smaller than the cofinality $\mathrm{cf}(\lambda )$ (see Definition 4.7.1.28).
Let $\{ T_{s} \} _{s \in S}$ be a collection of $\lambda $-small sets indexed by a set $S$ of cardinality $\leq \kappa $. Then the coproduct ${\coprod }_{s \in S} T_{s}$ is $\lambda $-small.
Proof. Assume first that condition $(1)$ is satisfied. Let $\{ T_{s} \} _{s \in S}$ be a collection of $\lambda $-small sets indexed by a set $S$ of cardinality $\leq \kappa $; we wish to show that the coproduct $T = {\coprod }_{s \in S} T_{s}$ is $\lambda $-small. Using Theorem 4.7.1.34, we can choose a well-ordering $\leq _{S}$ on the set $S$, and a well-ordering $\leq _{s}$ on the set $T_{s}$ for each $s \in S$. For elements $t \in T_{s}$ and $t' \in T_{s'}$, write $t \leq _{T} t'$ if either $s <_{S} s'$, or $s = s'$ and $t \leq _{s} t'$. Then $\leq _{T}$ is a well-ordering of the set $T$. If $T$ is not $\lambda $-small, then it has an initial segment of order type $\lambda $. Passing to subsets, we may assume without loss of generality that $T$ itself has order type $\lambda $. Moreover, we may assume without loss of generality that each of the sets $T_{s}$ is nonempty, and therefore contains a smallest element $t_{s}$. We consider two cases:
Suppose that $S$ contains a largest element $s$. In this case, we can write $T$ as the disjoint union of the initial segment $T' = {\coprod }_{s' < s} T_{s'}$ with the set $T_{s}$. Since $T_{s}$ is nonempty, $T'$ has order type smaller than $\lambda $, and is therefore $\lambda $-small. Applying Corollary 4.7.3.6, we deduce that $T = T' \coprod T_{s}$ is also $\lambda $-small.
Suppose that $S$ does not have a largest element. In this case, the construction $(s \in S) \mapsto (t_ s \in T)$ is a cofinal function from $S$ to $T$. It follows that the order type of $(S, \leq _{S} )$ is greater than or equal to the cofinality $\mathrm{cf}(T) = \mathrm{cf}(\lambda )$, contradicting assumption $(1)$.
We now prove the reverse implication. Assume that condition $(2)$ is satisfied. Choose a well-ordering $(S, \leq _{S} )$ of order type $\mathrm{cf}(\lambda )$ and a cofinal map $f: S \rightarrow \mathrm{Ord}_{< \lambda }$. If $\kappa \geq \mathrm{cf}(\lambda )$, then condition $(2)$ implies that the disjoint union ${\coprod }_{s \in S} \mathrm{Ord}_{ < f(s) }$ is $\lambda $-small. Since $f$ is cofinal, the tautological map ${\coprod }_{s \in S} \mathrm{Ord}_{ < f(s) } \rightarrow \mathrm{Ord}_{< \lambda }$ is surjective. It follows that $\mathrm{Ord}_{< \lambda }$ is $\lambda $-small, which is a contradiction. $\square$