Showing elements of an ultrapower have the form $j(f)([id]_{\mathcal{U}})$

Question

I am in a set theory class and we just started talking about ultrapowers. My professor mentioned that, if $\kappa$ is a measurable cardinal and $\mathcal{U}$ is a nonprincipal $\kappa$-complete ultrafilter on $\kappa$, then it is an easy consequence of Łoś's theorem that, for any function $f : \kappa \rightarrow V$, the equivalence class $[f]_{\mathcal{U}}$ is the same as $j(f)([id]_{\mathcal{U}})$, where $j$ is the canonical embedding. But when I tried to prove this, it wasn't easy for me! I know $j(f)([id]_{\mathcal{U}})$ must be an element of the ultrapower, and so I set $j(f)([id]_{\mathcal{U}})$ equal to $[g]$ for some $g : \kappa \rightarrow V$. Then I tried to show $\left\{\xi < \kappa : f(\xi) = g(\xi) \right\} \in \mathcal{U}$. If this was true, then by Los's theorem we would get $[f]_{\mathcal{U}} = [g]_{\mathcal{U}} = j(f)([id]_{\mathcal{U}})$. But I don't really understand the relationship between $g$ and $f$ and kept going in circles (one thing I tried to show was that $[id]_{\mathcal{U}}$ was in $j(\left\{\xi < \kappa : f(\xi) = g(\xi) \right\})$, and that definitely didn't seem to be going anywhere). Where am I going wrong?

Answer 1

As a newbie for ultrapower, handling elements of an ultrapower is confusing and not easy. I also experienced trouble to evaluate $j(f)([\mathsf{id}])$.

For me, viewing $j(f)([\mathsf{id}])$ componentwise was helpful. Imagine the function $g$ representing $j(f)([\mathsf{id}])$. What is the value of that function if we put $\alpha$ into $g$? Well, we may view $j(f)$ as a function from $\kappa$ to $V$ that always returns $f$. $\mathsf{Id}$ is just the identity function. By assumption, $[g]=j(f)([\mathsf{Id}])$. Now let us see the $\alpha$th section of this equality, then it would be $$g(\alpha)=(j(f)(\alpha))(\mathsf{Id}(\alpha))=f(\alpha).$$ Thus we may guess $g=f$.

Of course, the above argument is not an actual proof. The point is that functions in set theory are binary relations technically. Hence the assertion $f(x)=y$ is syntactic sugar for $(x,y)\in f$.

Thus what we have to prove is $([\mathsf{Id}],[f])\in j(f)$, which is equivalent to $$\{\alpha<\kappa \mid (\alpha,f(\alpha))\in f\}\in\mathcal{U},$$ but you can easily see that the above sentence holds.