I couldn't think of a better question title. I am trying to understand the proof of theorem 1.8 in this introduction to cardinals.
Theorem 1.8. Let $\kappa\in CARD$. Let $X=\bigcup_{\alpha<\kappa}X_\alpha$, where $|X_\alpha|\leq\kappa$ for all $\alpha<\kappa$. Then $|X|\leq \kappa$. Proof. Using AC, let $f$ be a function with domain $\kappa$ such that for all $\alpha<\kappa$, $f(\alpha):\kappa\to X_\alpha$ is a bijection. Define $F:\kappa\times\kappa\to X$ by $F(\alpha,\beta)=f(\alpha)(\beta)$. Clearly $F$ is onto, and the result follows by lemma 1.7. [which says $\kappa\cdot\kappa=\kappa$]
How can we "use AC" to obtain $f$? Is $f$ the choice function? Shouldn't it map each $\alpha$ to a member of $\alpha$?
Also, how can we possibly ensure there is such a bijection for each $\alpha$ – we only know $|X_\alpha|\leq \kappa$, ie $|X_\alpha|<\kappa$ is possible! In that case how can there be a bijection?
First note that we only need $f(\alpha)\colon \kappa\to X_\alpha$ to be onto (after all the final $F$ is also only needed onto). So I suppose "bijection" is just a typo.
We are given for each $\alpha$ that $|X_\alpha|\le\kappa$, i.e., that the set $S_\alpha$ of surjective maps $\kappa\to X_\alpha$ is $\ne\emptyset$. We use AC to pick an element $f(\alpha)\in S_\alpha$ for each $\alpha$.