Proof.
Let \(X\in \mathrm{Alg}_{\mathbb E_1}(\mathcal C)\). Since \(Z_1(A)\) is the centralizer \(\mathfrak{Z}(\mathrm{id}_A)\) of the morphism \(\mathrm{id}_A\) in \(\mathrm{Alg}_{\mathbb E_1}(\mathcal C)\) and the unit of \(\mathrm{Alg}_{\mathbb E_1}(\mathcal C)\) is initial, the universal property of the centralizer implies that the map \(\mathrm{Hom}_{\mathrm{Alg}_{\mathbb E_1}(\mathcal C) }(X, Z_1(A)) \rightarrow\mathrm{Hom}_{\mathrm{Alg}_{\mathbb E_1}(\mathcal C)}(X, A)\) is equivalent to the composite \[\mathrm{Hom}_{\mathrm{Alg}_{\mathbb E_1}(\mathcal C)}(X \otimes A, A) \times_{\mathrm{Hom}_{\mathrm{Alg}_{\mathbb E_1}(\mathcal C)}(A,A)} \{ \mathrm{id}_A\} \rightarrow\mathrm{Hom}_{\mathrm{Alg}_{\mathbb E_1}(\mathcal C)}(X \otimes A, A) \rightarrow\mathrm{Hom}_{\mathrm{Alg}_{\mathbb E_1}(\mathcal C)}(X,A).\] By definition of the \(\infty\)-operad \(\nabla_2\) in lemma 7.2.3, the fiber of this map at an \(f\in \mathrm{Hom}_{\mathrm{Alg}_{\mathbb E_1}(\mathcal C)}(X,A)\) is precisely the space of lifts of the operad map 
classified by the span of \(\mathbb E_1\)-morphisms \(X \xrightarrow{f}A \xleftarrow{\mathrm{id}_A}A\), to an operad map \(\nabla_2 \otimes \mathbb E_0 \rightarrow\mathrm{Alg}_{\mathbb E_1}(\mathcal C)\). Equivalently, this is the space of lift 
By assumption, \(\mathcal C\) is a symmetric monoidal \((2,1)\)-category, hence a \(2\)-operad and hence the operad map \(\mathcal C\rightarrow*\) is \(2\)-faithful. Since the left vertical operad map is \(0\)-surjective by proposition 7.6.1, it follows from corollary 7.5.5 that this space of lifts is \(0\)-truncated. ◻