$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$
Corollary 11.9.1.6. Let $\operatorname{\mathcal{C}}$ be an $\infty $-category and let $f$ be a morphism of $\operatorname{\mathcal{C}}$. Then $f$ is an isomorphism if and only if it is universal with respect to the balanced coupling
\[ \lambda : \operatorname{Cospan}^{L,R}( \operatorname{Fun}(\Delta ^1, \operatorname{\mathcal{C}}) ) \rightarrow \operatorname{Fun}^{ \mathrm{iso}, \mathrm{all} }( \operatorname{\mathcal{C}}) \times \operatorname{Fun}^{ \mathrm{all}, \mathrm{iso} }( \operatorname{\mathcal{C}}) \]
of Proposition 11.9.1.1 (where we abuse notation by identifying $f$ with an object of the $\infty $-category $\operatorname{Cospan}^{L,R}( \operatorname{Fun}(\Delta ^1, \operatorname{\mathcal{C}}) )$).
Proof.
Let us abuse notation further by identifying $f$ with an object of the twisted arrow $\infty $-category $\operatorname{Tw}(\operatorname{\mathcal{C}})$, so that the comparison functor $\Xi : \operatorname{Tw}(\operatorname{\mathcal{C}}) \rightarrow \operatorname{Cospan}^{L,R}( \operatorname{Fun}(\Delta ^1, \operatorname{\mathcal{C}}) )$ of Construction 11.9.1.3 satisfies $\Xi (f) = f$ (Remark 11.9.1.4). By virtue of Proposition 11.9.1.5 and Remark 8.3.2.8, we are reduced to showing that $f$ is an isomorphism if and only if it is universal with respect to the twisted arrow coupling $\operatorname{Tw}(\operatorname{\mathcal{C}}) \rightarrow \operatorname{\mathcal{C}}^{\operatorname{op}} \times \operatorname{\mathcal{C}}$, which follows from Example 8.2.1.5.
$\square$