Corollary 4.4.3.8. Let $q: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be a morphism of simplicial sets, where $\operatorname{\mathcal{D}}$ is a Kan complex. The following conditions are equivalent:
The morphism $q$ is a Kan fibration.
The morphism $q$ is a left fibration.
The morphism $q$ is a right fibration.
The morphism $q$ is a conservative isofibration of $\infty $-categories.
Proof. The implications $(1) \Rightarrow (2) \Rightarrow (4)$ and $(1) \Rightarrow (3) \Rightarrow (4)$ follow from Example 4.2.1.5, Proposition 4.4.2.11, and Example 4.4.1.11 (and do not require the assumption that $\operatorname{\mathcal{D}}$ is a Kan complex). We will complete the proof by showing that $(4) \Rightarrow (1)$. Our assumption that $\operatorname{\mathcal{D}}$ is a Kan complex guarantees that every morphism in $\operatorname{\mathcal{D}}$ is an isomorphism. Since $q$ is conservative, it follows that every morphism in $\operatorname{\mathcal{C}}$ is an isomorphism. We can therefore identify $q$ with the induced map $q^{\simeq }: \operatorname{\mathcal{C}}^{\simeq } \rightarrow \operatorname{\mathcal{D}}^{\simeq }$, which is Kan fibration by virtue of Proposition 4.4.3.7. $\square$