Equivalent formulations of Woodin cardinals Consider the following properties for a given ordinal $\delta$:


*

*For all $f \colon \delta \to \delta$ there is an elementary embedding
$$
j \colon V \to M, M \text{ transitive}
$$
such that $f " \mathrm{crit}(j) \subseteq \mathrm{crit}(j)$ and $V_{j(f)(\mathrm{crit}(j))} \subseteq M$.

*For all $A \subseteq V_{\delta}$ there is some $\kappa < \delta$ that is $\gamma$-strong for $A$ for all $\gamma < \delta$, i.e. for every $\gamma < \delta$ there is an elementary embedding
$$
j_{\gamma} \colon V \to M_{\gamma}, M_{\gamma} \text{ transitive }
$$
such that $\kappa = \mathrm{crit}(j)$, $j(\kappa) > \gamma$, $V_{\kappa + \gamma} \subseteq M_{\gamma}$ and $A \cap V_{\kappa + \gamma} = j(A) \cap V_{\kappa + \gamma}$.


In "The Higher Infinite" 26.14, Kanamori claims that 1. and 2. are equivalent. However, it seems that a part of the proof for this is missing in my edition: 
Question. Assuming 2., how do we obtain (for a fixed $f$) an elementary embedding $j \colon V \to M$ with $V_{j(f)(\mathrm{crit}(j))}\subseteq M$?
 A: Below is the argument in Kanamori's book with more details.
Suppoose $\delta $ is such that for all $ A \subseteq V_{\delta}$ there is $\alpha < \delta $ such that $\alpha$ is $\gamma$-strong for $A$ for every $\gamma < \delta$. Then for all $f \in \delta^{\delta}$ there is $\alpha < \delta$ such that $f''\alpha \subseteq \alpha$ and $j_{E}:V \rightarrow M$ an elementary embedding such that $crit(j_{E}) = \alpha$ and $V_{j_{E}(f)(\alpha)} \subseteq M$ where $j_{E}$ is the elementary embedding given by an extender $E \in V_{\delta}$ 


*

*For all $g \in \delta^{\delta}$ if $\alpha$ is $\gamma$-strong for $g$ for all $\gamma < \delta$, then $g'' \alpha \subseteq \alpha$.


Let $ g \in \delta^{\delta}$, $ A = g $ and $\alpha$ such that $\alpha$ is $\gamma$-strong for $A$ for all $\gamma < \delta$. Fix $\xi < \alpha$ and $j$ an elementary embedding such that 
\begin{equation} (*) \ m_{g(\xi)} := max \{ \xi , g(\xi) + 1000 \} < j(\alpha),
\end{equation} which witness that $\alpha$ is $\gamma$-strong for $A$ where $\gamma = m_{g(\xi)}$. 
We want to see that $g''\alpha \subseteq \alpha$.
We have  \begin{gather*}  \langle (\beta,g(\beta)) \ | \ \beta < \delta \rangle \cap V_{\alpha + \gamma} = j(g) \cap   V_{\alpha + \gamma} 
\end{gather*} 
Notice that 
\begin{gather*} (\xi,g(\xi)) \in g \cap V_{\alpha + \gamma},
\end{gather*}  
\begin{gather*} (\xi,g(\xi) \in  g \cap V_{\alpha + \gamma}  =  j(g) \cap   V_{\alpha + \gamma} \subseteq j(g) 
\end{gather*} 
and 
\begin{gather}(**) \  j(g)(\xi) = g(\xi) < j(\alpha)
\end{gather} 
where the $<$ follows from the choice of $j$ (see (*)), but by elementarity 
\begin{gather*}  j(g(\xi)) = j(g)(j(\xi)) = j(g)(\xi) < j(\alpha) \longrightarrow  g(\xi) < \alpha 
\end{gather*} 
where the second equality follows from $\xi < crit(j) $ and $< $ follows from (**). 
Since $\xi$ was arbitrary below $\alpha$ it follows that $g''\alpha \subseteq \alpha$.
This  shows that for all $g \in \delta^{\delta}$ if $\alpha$ is $\gamma$-strong for $g$ for all $\gamma < \delta$, then $g'' \alpha \subseteq \alpha$.
To verify that $\delta$ is regular  suppose not and let $f \in \delta^{\delta}$ such that $f(0) > cf(\delta)$ and $f''cf(\delta)$ is cofinal in $\delta$. Then if $\alpha \leq \delta$ and $f''\alpha \subseteq \alpha$ it follows that $\alpha > cf(\delta)$, then $f'' cf(\delta) \subseteq \alpha $ so $\alpha = \delta$, which contradicts what we have just shown for $g$ above.



*$\delta$ is a inaccessible cardinal (we also show that it is a limit of measurable cardinals)


Fix $\eta < \delta$ and take $f_{\eta} \in \delta^{\delta}$ such that $f_{\eta}(0) > \eta$. Apply 1  for $g = f_{\eta}$, then we find $\alpha < \delta $ measurable such that $f''\alpha \subseteq \alpha $, so $0 \in \alpha $ and $\eta = f_{\eta}(0) < \alpha $.  Thus for every $\eta < \delta $ there is $\alpha$ such that $\eta < \alpha < \delta$ a measurable cardinal. This shows that $\delta $ is inaccessible and a limit of measurable cardinals.  



*Given $f \in \delta^{\delta}$ there is $j:V \longrightarrow M $ such that $crit(j) = \alpha < \delta $ and $V_{j(f)(\alpha)} \subseteq M $. 


We do a small variation on the argument of 1. Let $\alpha$ be $\gamma $-strong for $f$ for all $\gamma < \delta$. Fix $ \gamma  > max\{\alpha,f(\alpha) + 1000\}$ and let $j$ witness that $\alpha$ is $\gamma$-strong for $f$. 
Then 
\begin{gather*} (\alpha,f(\alpha)) \in f \cap V_{\alpha + \gamma} = j(f) \cap V_{\alpha + \gamma} \subseteq j(f)
\end{gather*} 
so $j(f)(\alpha)=f(\alpha)$ and $\gamma \geq f(\alpha)+1000 > f(\alpha)$ so $V_{f(\alpha)} \subseteq V_{\gamma} \subseteq M$. 
\end{proof}
It is straightforward to derived the extenders advertised above.

The argument gives the following: Let $f \in \delta^{\delta}$ and $\eta < \delta$, then there is $j:V \rightarrow M$ such that $j(f)(\eta) = f(\eta) $ and $V_{\eta} \subseteq M$. 
For that we let $\alpha$ be $\gamma$-strong for $f$ for all $ \gamma < \delta$. Let $ \gamma > max\{ \eta, f(\eta)+1000\} $ then  $V_{\alpha + \gamma} \subseteq M$ and
\begin{gather*} (\eta,f(\eta)) \in f \cap V_{\alpha + \gamma}  = j(f) \cap V_{\alpha + \gamma} \subseteq j(f)
\end{gather*}  which implies $j(f)(\eta) = f(\eta)$.
