Lemma 7.1.4.26. Let $\operatorname{\mathcal{C}}$ be an $\infty $-category and let $\operatorname{\mathcal{C}}' \subseteq \operatorname{\mathcal{C}}$ be a reflective subcategory. If $\operatorname{\mathcal{C}}$ contains a final object $X$, then $\operatorname{\mathcal{C}}'$ contains an object which is isomorphic to $X$. In particular, if $\operatorname{\mathcal{C}}'$ is replete, then it contains every final object of $\operatorname{\mathcal{C}}$.
$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$
Proof. Choose a morphism $f: X \rightarrow Y$ in $\operatorname{\mathcal{C}}$ which exhibits $Y$ as a $\operatorname{\mathcal{C}}'$-reflection of $X$ (see Definition 6.2.2.1). Since $X$ is a final object of $\operatorname{\mathcal{C}}$, we can choose a morphism $g: Y \rightarrow X$. We will complete the proof by showing that $g$ is a homotopy inverse to $f$: that is, the homotopy classes $[f]$ and $[g]$ are inverses of one another in the homotopy category $\mathrm{h} \mathit{\operatorname{\mathcal{C}}}$. Since $X$ is a final object of $\operatorname{\mathcal{C}}$, the equality $[g] \circ [f] = [ \operatorname{id}_ X ]$ is automatic. We wish to prove the equality $[f] \circ [g] = [ \operatorname{id}_ Y ]$. Since $f$ exhibits $Y$ as a $\operatorname{\mathcal{C}}'$-reflection of $X$, precomposition with $[f]$ induces a bijection $\operatorname{Hom}_{\mathrm{h} \mathit{\operatorname{\mathcal{C}}}}( Y, Y ) \rightarrow \operatorname{Hom}_{ \mathrm{h} \mathit{\operatorname{\mathcal{C}}}}( X, Y)$. The desired result now follows from the calculation
\[ ([f] \circ [g]) \circ [f] = [f] \circ ([g] \circ [f] ) = [f] \circ [ \operatorname{id}_ X ] = [f] = [ \operatorname{id}_{Y} ] \circ [f]. \]
$\square$