Proof. First assume \(X\) is \(P\)-divisible. To see that \(X\) is \(\S [P^{-1}]\)-local, consider an \(\S [P^{-1}]\)-acyclic spectrum \(Y\). We must show that \(\hom (Y,X) = 0\). This follows from the following sequence of isomorphisms: \[ \hom (Y,X) \cong \hom (Y[P^{-1}],X) \cong \hom (0,X) \cong 0; \] Here the first isomorphism follows from Proposition 7.2.7, and the second from the observation that \(Y[P^{-1}] \cong Y\otimes \S [P^{-1}]\).
Conversely, assume that \(X\) is \(\S [P^{-1}]\)-local. To show that \(p\colon X \to X\) is an isomorphism, it suffices by Lemma 7.1.6 to show that the map \(p\colon X[P^{-1}] = X \otimes \S [P^{-1}] \to X \otimes \S [P^{-1}] = X[P^{-1}]\) is an isomorphism, which is the content of Corollary 7.2.5. □
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