Lemma 3.7.1. Suppose that $T = \mathop{\mathrm{colim}}\nolimits _{\alpha < \beta } T_\alpha $ is a colimit of sets indexed by ordinals less than a given ordinal $\beta $. Suppose that $\varphi : S \to T$ is a map of sets. Then $\varphi $ lifts to a map into $T_\alpha $ for some $\alpha < \beta $ provided that $\beta $ is not a limit of ordinals indexed by $S$, in other words, if $\beta $ is an ordinal with $\text{cf}(\beta ) > |S|$.

## 3.7 Cofinality

A *cofinal subset* $S$ of a well-ordered set $T$ is a subset $S \subset T$ such that $\forall t \in T \exists s\in S (t \leq s)$. Note that a subset of a well-ordered set is a well-ordered set (with induced ordering). Given an ordinal $\alpha $, the *cofinality* $\text{cf}(\alpha )$ of $\alpha $ is the least ordinal $\beta $ which occurs as the order type of some cofinal subset of $\alpha $. The cofinality of an ordinal is always a cardinal. Hence alternatively we can define the cofinality of $\alpha $ as the least cardinality of a cofinal subset of $\alpha $.

**Proof.**
For each element $s \in S$ pick a $\alpha _ s < \beta $ and an element $t_ s \in T_{\alpha _ s}$ which maps to $\varphi (s)$ in $T$. By assumption $\alpha = \sup _{s \in S} \alpha _ s$ is strictly smaller than $\beta $. Hence the map $\varphi _\alpha : S \to T_\alpha $ which assigns to $s$ the image of $t_ s$ in $T_\alpha $ is a solution.
$\square$

The following is essentially Grothendieck's argument for the existence of ordinals with arbitrarily large cofinality which he used to prove the existence of enough injectives in certain abelian categories, see [Tohoku].

Proposition 3.7.2. Let $\kappa $ be a cardinal. Then there exists an ordinal whose cofinality is bigger than $\kappa $.

**Proof.**
If $\kappa $ is finite, then $\omega = \text{cf}(\omega )$ works. Let us thus assume that $\kappa $ is infinite. Consider the smallest ordinal $\alpha $ whose cardinality is strictly greater than $\kappa $. We claim that $\text{cf}(\alpha ) > \kappa $. Note that $\alpha $ is a limit ordinal, since if $\alpha = \beta + 1$, then $|\alpha | = |\beta |$ (because $\alpha $ and $\beta $ are infinite) and this contradicts the minimality of $\alpha $. (Of course $\alpha $ is also a cardinal, but we do not need this.) To get a contradiction suppose $S \subset \alpha $ is a cofinal subset with $|S| \leq \kappa $. For $\beta \in S$, i.e., $\beta < \alpha $, we have $|\beta | \leq \kappa $ by minimality of $\alpha $. As $\alpha $ is a limit ordinal and $S$ cofinal in $\alpha $ we obtain $\alpha = \bigcup _{\beta \in S} \beta $. Hence $|\alpha | \leq |S| \otimes \kappa \leq \kappa \otimes \kappa \leq \kappa $ which is a contradiction with our choice of $\alpha $.
$\square$

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