[proofplan] Let $\delta = \operatorname{cf}(\alpha)$. By definition, $\delta$ is the least ordinal admitting a cofinal map into $\alpha$. We show that no smaller ordinal can be equipotent to $\delta$: otherwise a bijection from that smaller ordinal onto $\delta$ would reindex a cofinal map $\delta \to \alpha$ and produce a cofinal map from a strictly smaller domain, contradicting the minimality in the definition of cofinality. [/proofplan] [step:Choose a least cofinal domain for $\alpha$] Let $\delta := \operatorname{cf}(\alpha)$. By the definition of cofinality, $\delta$ is an ordinal such that there exists a map $f: \delta \to \alpha$ whose range is cofinal in $\alpha$, and $\delta$ is least with this property. Thus, for every $\beta < \alpha$, there exists $\xi < \delta$ such that $\beta \leq f(\xi)$. [/step] [step:Show that an equipotent smaller ordinal would reindex the cofinal map] Suppose, toward a contradiction, that $\delta$ is not a cardinal. Since cardinals are initial ordinals, there exists an ordinal $\lambda < \delta$ and a bijection $b: \lambda \to \delta$. Define the composite map $h: \lambda \to \alpha$ by \begin{align*} h(\eta) := f(b(\eta)). \end{align*} We claim that the range of $h$ is cofinal in $\alpha$. Let $\beta < \alpha$. Since the range of $f$ is cofinal in $\alpha$, there exists $\xi < \delta$ such that $\beta \leq f(\xi)$. Since $b: \lambda \to \delta$ is surjective, there exists $\eta < \lambda$ such that $b(\eta) = \xi$. Therefore \begin{align*} \beta \leq f(\xi) = f(b(\eta)) = h(\eta). \end{align*} Hence $h: \lambda \to \alpha$ has cofinal range. [guided] Assume that $\delta$ is not an initial ordinal. By the definition of an initial ordinal, this means that some strictly smaller ordinal has the same cardinality as $\delta$. Therefore there is an ordinal $\lambda < \delta$ and a bijection $b: \lambda \to \delta$. The purpose of the bijection is to transfer the cofinal map from the domain $\delta$ to the smaller domain $\lambda$. Define $h: \lambda \to \alpha$ by \begin{align*} h(\eta) := f(b(\eta)). \end{align*} This is a well-defined map into $\alpha$ because $b(\eta) \in \delta$ for every $\eta < \lambda$, and $f$ is defined on all of $\delta$. We verify that $h$ has cofinal range in $\alpha$. Let $\beta < \alpha$. Since $f: \delta \to \alpha$ has cofinal range, there exists $\xi < \delta$ such that \begin{align*} \beta \leq f(\xi). \end{align*} Because $b$ is surjective onto $\delta$, this particular $\xi$ has a preimage under $b$: there exists $\eta < \lambda$ such that $b(\eta) = \xi$. Substituting this equality gives \begin{align*} \beta \leq f(\xi) = f(b(\eta)) = h(\eta). \end{align*} Since this works for every $\beta < \alpha$, the range of $h$ is cofinal in $\alpha$. [/guided] [/step] [step:Contradict the minimality of $\operatorname{cf}(\alpha)$] The map $h: \lambda \to \alpha$ has cofinal range, while $\lambda < \delta = \operatorname{cf}(\alpha)$. This contradicts the defining minimality of $\delta$ as the least ordinal admitting a cofinal map into $\alpha$. Therefore no ordinal $\lambda < \delta$ is equipotent to $\delta$. Hence $\delta$ is the initial ordinal of its cardinality, so $\delta$ is a cardinal. Since $\delta = \operatorname{cf}(\alpha)$, this proves that $\operatorname{cf}(\alpha)$ is a cardinal. [/step]