Let $\kappa$ be measurable and $U$ the ultrafilter. Consider the ultrapower $\pi: V^{\kappa}/U \cong M$, where $\pi$ is the transitive collapse. We know that $\kappa \in M$, so there is some $f:\kappa \to V$ such that $[f]_U=\kappa$.
But I read in a proof that we can pick $f$ to have codomain $\kappa$ instead of $V$. But this seems wrong to me, since if $\{\alpha < \kappa : f(\alpha)<\kappa\} \in U$, then by Łoś we would have $M\models [f]_U<\kappa$, right? This means such an $f$ can't have codomain $\kappa$.
What am I missing?
Actually, Łoś would tell you that for $f$ with codomain $\kappa$ you get $[f]_U<[c_\kappa]_U$, where $c_\kappa$ is the constant function with value $\kappa$. Or in other words, $[f]_U<j(\kappa)$, which is perfectly reasonable. In fact, the argument can be reversed, and any element of the ultrapower below $j(\kappa)$ can be represented by a function with codomain $\kappa$.