Corollary 9.2.2.19. Let $\operatorname{\mathcal{C}}$ be an $\infty $-category and let $W$ be a collection of morphisms of $\operatorname{\mathcal{C}}$ which is closed under isomorphism. Then a morphism $f$ of $\operatorname{\mathcal{C}}$ belongs to the transfinite closure of $W$ if and only if $f$ is either an isomorphism or a transfinite composition of morphisms of $W$.
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Proof. Assume that $f$ belongs to the transfinite closure of $W$; we will show that $f$ is either an isomorphism or a transfinite composition of morphisms of $W$ (the converse is clear, since the transfinite closure of $W$ contains all isomorphisms: see Remark 9.2.2.8). Let $W^{+}$ be the union of $W$ with the collection of all identity morphisms of $\operatorname{\mathcal{C}}$. Applying Proposition 9.2.2.16, we see that $f$ is a transfinite composition of morphisms of $W^{+}$. If $f$ is not an isomorphism, then Proposition 9.2.2.18 guarantees that $f$ is a transfinite composition of morphisms which belong to $W^{+}$ and are not isomorphism, and therefore belong to $W$. $\square$