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Gabe Goldberg
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Suppose $\delta$ is an ordinal. We first note that $\delta$ is an $I_1$-tower cardinal if and only if it has the following superficially weaker property: for all $X\subseteq V_\delta$ there is some $\kappa < \delta$ such that for arbitrarily large $\lambda < \delta$, there is an elementary embedding $j:V_{\lambda+1}\to V_{\lambda+1}$ with critical point $\kappa$ fixing $X\cap V_\lambda$. (This is because if there is an embedding $j:V_{\lambda+1}\to V_{\lambda+1}$ with critical point $\kappa$, there is an embedding $j':V_{\lambda+1}\to V_{\lambda+1}$ with critical point $\kappa$ that sends $\kappa$ arbitrarily high below $\lambda$ and fixes anything that $j$ fixes: just let $j' = j\circ j \circ \cdots\circ j$.)

Suppose $\delta$ is the critical point of an embedding $j: L_1(V_{\lambda+1})\to L_1(V_{\lambda+1})$. We show $\delta$ is an $I_1$-tower cardinal. Fix $X\subseteq V_\delta$. Let $j_{0\omega}: V_{\lambda+1}\to M_\omega$ denote the $\omega$-th iterate of $j$. Let $X_* = j_{0\omega}(X)$. A standard fact is that $j$ fixes $X_*$: $$j(X_*) = j(j_{0\omega}(X)) = j_{1\omega}(j(X)) = j_{0\omega}(X) = X_*$$ We make the following claim: there exist arbitrarily large $\bar \lambda < \lambda$ such that there is an elementary embedding $\bar j : V_{\bar \lambda+1}\to V_{\bar \lambda+1}$ with critical point $\delta$ fixing $X_*\cap V_{\bar \lambda}$. This is proved by Laver's inverse limit reflection technique, but assume it is true for the moment and let us conclude the proof. Since $V_\lambda \subseteq M_\omega$, the claim is absolute to $M_\omega$. Pulling the claim back from $M_\omega$ to $V_{\lambda+1}$ through $j_{0\omega}$, and noting that $j_{0\omega}(\delta) = \lambda$ and $j_{0\omega}(X) = X_*$, there is some $\kappa < \delta$ such that for arbitrarily large $\bar \lambda < \delta$, there is an elementary embedding $\bar j: V_{\bar \lambda +1 }\to V_{\bar \lambda +1 }$ with critical point $\kappa$ fixing $X\cap V_{\bar \lambda}$. This shows that $\delta$ has the superficially weakened $I_1$-tower property with respect to $X$, as desired.

We now prove the claim. Fix an arbitrary $\alpha < \lambda$, and assume for ease of notation that $\alpha > \delta$. We must find an ordinal $\bar \lambda\in (\alpha,\lambda]$ and an elementary embedding $\bar j : V_{\bar \lambda+1}\to V_{\bar \lambda+1}$ fixing $X_*\cap V_{\bar \lambda}$ with critical point $\delta$. Inverse limit reflection yields an ordinal $\bar \lambda\in (\alpha, \lambda]$ and an elementary embedding $J:V_{\bar \lambda + 1}\to V_{\lambda+1}$ with critical point above $\alpha$ sending $X_*\cap V_{\bar \lambda}$ to $X_*$ with $j\restriction V_\lambda$ in the range of $J$. The only part of this that is not a clause in the standard inverse limit lemma is that we can find $J$ sending $X_*\cap V_{\bar \lambda}$ to $X_*$, so we indicate how this part is achieved. We first iterate $j$ finitely many times to obtain an embedding $j_n : L_1(V_{\lambda+1})\to L_1(V_{\lambda+1})$ with critical point above $\alpha$, in order to ensure that the critical point of $J$ is above $\alpha$, though we will ignore this part of the construction. Of course $j_n$ fixes $X_*$ since $j$ does. We build $J$ as an inverse limit of embeddings $\langle k_i: i < \omega\rangle$, choosing the component embeddings $k_i:V_{\lambda+1}\to V_{\lambda+1}$ to be square roots of $j$ that also fix $X_*$, which is possible by the square root lemma. (To obtain square roots, we use the assumptiom that $j$ is stronger than just an $I_1$-embedding.) Letting $J = k_0\circ k_1\circ k_2\circ \cdots$, we have $k_0\circ\cdots \circ k_n(X_*\cap V_{\text{crt}(k_n)}) = X_*\cap V_{k_0\circ\cdots \circ k_n(\text{crt}(k_n))}$. Hence $J(X_*\cap V_{\bar \lambda}) = X_*$ by the definition of an inverse limit. (For the rest of the details of this construction, see Laver's paper "Implications between strong large cardinal axioms.") Now let $\bar j = J^{-1}(j\restriction V_\lambda)$. Since $J:V_{\bar \lambda+1}\to V_{\lambda+1}$ is elementary, $\bar j$ extends to an elementary embedding $V_{\bar \lambda + 1} \to V_{\bar \lambda + 1}$ with critical point $J^{-1}(\delta) = \delta$ fixing $J^{-1}(X_*) = X_*\cap V_{\bar \lambda}$. This proves the claim.

Gabe Goldberg
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