Theorem 7.3.5. For a spectrum \(X\), the map \[ X = \hom (\S ,X) \to \hom (\S /p^{\infty }[-1],X) \] induced by the map \(\S /p^{\infty }[-1] \to \S \) exhibits its target as a \(p\)-completion of \(X\), i.e. \[ X^{\wedge }_p \iso \hom (\S /p^{\infty }[-1],X) \iso \hom (\S /p^{\infty },X)[1]. \]
Proof. We have to show that \(\hom (\S /p^{\infty }[-1],X)\) is \(p\)-complete and that the map \(X \to \hom (\S /p^{\infty }[-1],X)\) is an \(\S /p\)-equivalence. For the latter, we may equivalently show that its fiber is \(\S /p\)-acyclic, or equivalently that its fiber is \(\S [\frac {1}{p}]\)-local by Observation 7.3.2. Because of the exact sequence \(\S /p^{\infty }[-1] \to \S \to \S [\frac {1}{p}]\), this fiber is isomorphic to \(\hom (\S [\frac {1}{p}],X)\). But this is clearly \(\S [\frac {1}{p}]\)-local: for any \(\S [\frac {1}{p}]\)-acyclic spectrum \(Y\) we have \(\hom (Y,\hom (\S [\frac {1}{p}],X)) \simeq \hom (Y \otimes \S [\frac {1}{p}],X) \simeq \hom (0,X) = 0\).
We will now show that \(\hom (\S /p^{\infty }[-1],X)\) is \(p\)-complete, i.e. \(\S /p\)-local. To this end, let \(A\) be an \(\S /p\)-acyclic spectrum, i.e. the map \(p\colon A \to A\) is an isomorphism. We need to show that \(0 = \hom (A,\hom (\S /p^{\infty }[-1],X)) \simeq \hom (A \otimes \S /p^{\infty }[-1],X)\). For this, it will suffice to show that \(A \otimes \S /p^{\infty } = 0\). Since we may write \(\S /p^{\infty }\) as a colimit of \(\S /p^n\), we may similarly write \[ A \otimes \S /p^{\infty } \simeq \colim _n(A \otimes \S /p^n) \simeq \colim _n(A/p^n). \] But as \(A\) is \(\S /p\)-acyclic, the map \(p\colon A\to A\) and hence also \(p^n\colon A \to A\) is an isomorphism, and so \(A/p^n = 0\). This shows that \(A \otimes \S /p^{\infty }\simeq 0\), as desired. □
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