Why isn't the cofinality of $\lambda^\kappa$ equal to $\kappa$, in contradiction to König's theorem?

Question

I came across the definition of cofinality and there is something I don't understand : König's theorem tells us that $\operatorname{cof}(\lambda^\kappa)>\kappa$ but since $\lambda^\kappa:=\bigcup\limits_{\gamma<\kappa}\lambda^{\gamma}$, the map \begin{align} \kappa &\to \lambda^{\kappa}\\ \gamma &\mapsto \lambda^{\gamma} \end{align} should be cofinal no? So $\operatorname{Cof}(\lambda^{\kappa})\le \kappa$ no? What did I get wrong here?

Answer 1

$\lambda^\kappa:=\bigcup\limits_{\gamma<\kappa}\lambda^{\gamma}$ holds for ordinals (if $\kappa$ is a limit, see here e.g.), but the statement for the cofinality is a statement about cardinals.

You would not say that $\bigcup_{n < \aleph_0} \aleph_0^n = \aleph_0^{\aleph_0}$ would you?

Answer 2

It is absolutely not true that $\lambda^\kappa=\bigcup_{\gamma<\kappa}\lambda^\gamma$. Just take $\kappa=\omega$ to understand how wrong that may be.

What you describe is $\lambda^{<\kappa}$, which indeed has a cofinal sequence of length $\kappa$, although it might very well be that $\lambda^\gamma$ is constant. Cardinal exponentiation is kinda tricky this way.