[00DS]
Proof.
Part ([00DQ]) follows directly from [Lur17, Prop. 2.3.1.9]. For part ([00DR]), note that for an \(\infty\)-operad \(\mathcal P\), the fiber of \(\mathrm{Hom}_{\mathrm{Op}}(\mathbb E_0, \mathcal P) \rightarrow\mathrm{Hom}_{\mathrm{Op}}(\mathrm{Triv}, \mathcal P) = \underline{ \mathcal P}^{\simeq}\) at a color \(X\in \underline{\mathcal P}\) is the \(0\)-ary mapping space \(\mathrm{Mul}_{\mathcal P}(\emptyset, X)\) and hence that \(\mathcal P\) is unital if and only if \(\mathrm{Hom}_{\mathrm{Op}}(\mathbb E_0, \mathcal P) \rightarrow\mathrm{Hom}_{\mathrm{Op}}(\mathrm{Triv}, \mathcal P)\) is an isomorphism. In particular, evaluating at \(\mathcal P= \mathrm{Alg}_{\mathbb E_0}(\mathcal O)\) for an \(\infty\)-operad \(\mathcal O\), and using that \(\mathrm{Hom}_{\mathrm{Op}}(-, \mathrm{Alg}_{\mathbb E_0}(\mathcal O)) \simeq \mathrm{Hom}_{\mathrm{Op}}(- \otimes \mathbb E_0, \mathcal O)\) and part ([00DQ]), it follows that \(\mathrm{Alg}_{\mathbb E_0}(\mathcal O)\) is unital. If \(\mathcal O\) is moreover unital, let \(\mathcal Q\) be a unital \(\infty\)-operad and consider the map \(\mathrm{Hom}_{\mathrm{Op}^{\mathrm{un}}}(\mathcal Q, \mathrm{Alg}_{\mathbb E_0}(\mathcal O))\rightarrow\mathrm{Hom}_{\mathrm{Op}^{\mathrm{un}}}(\mathcal Q, \mathcal O)\) which is equivalent to \(\mathrm{Hom}_{\mathrm{Op}^{\mathrm{un}}}(\mathcal Q\otimes \mathbb E_0, \mathcal O) \rightarrow\mathrm{Hom}_{\mathrm{Op}^{\mathrm{un}}}(\mathcal Q, \mathcal O)\). Since \(\mathcal Q\) is unital, this is an isomorphism by part ([00DQ]). This completes the proof of part ([00DR]). ◻