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Proposition 5.6.6.20. Let $U: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{C}}$ be a left fibration of $\infty $-categories and let $\widetilde{X} \in \operatorname{\mathcal{D}}$ be an object having image $X = U( \widetilde{X} )$. The following conditions are equivalent:

$(1)$

There exists an equivalence $F: \operatorname{\mathcal{C}}_{X/} \rightarrow \operatorname{\mathcal{D}}$ of left fibrations over $\operatorname{\mathcal{C}}$ satisfying $F( \operatorname{id}_ X) = \widetilde{X}$.

$(2)$

The object $\widetilde{X} \in \operatorname{\mathcal{D}}$ is initial (Definition 4.6.6.1).

$(3)$

For every left fibration $V: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$, evaluation on the object $\widetilde{X}$ induces a trivial Kan fibration $\operatorname{Fun}_{ / \operatorname{\mathcal{C}}}( \operatorname{\mathcal{D}}, \operatorname{\mathcal{E}}) \rightarrow \{ X\} \times _{\operatorname{\mathcal{C}}} \operatorname{\mathcal{E}}$.

$(4)$

For every left fibration $V: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$, evaluation on the object $\widetilde{X}$ induces a bijection

\[ \pi _0( \operatorname{Fun}_{ / \operatorname{\mathcal{C}}}( \operatorname{\mathcal{D}}, \operatorname{\mathcal{E}}) ) \rightarrow \pi _0( \{ X\} \times _{\operatorname{\mathcal{C}}} \operatorname{\mathcal{E}}). \]

Proof. If $F: \operatorname{\mathcal{C}}_{X/} \rightarrow \operatorname{\mathcal{D}}$ is an equivalence of left fibrations over $\operatorname{\mathcal{C}}$, then it is an equivalence of $\infty $-categories (Proposition 5.1.6.5). Since $\operatorname{id}_{X}: X \rightarrow X$ is initial when regarded as an object of the $\infty $-category $\operatorname{\mathcal{C}}_{X/}$ (Proposition 4.6.6.23), Corollary 4.6.6.21 guarantees that $\widetilde{X}$ is an initial object of $\operatorname{\mathcal{D}}$. This proves the implication $(1) \Rightarrow (2)$. The implication $(2) \Rightarrow (3)$ follows by combining Corollary 4.6.6.25 with Proposition 4.2.5.4, and the implication $(3) \Rightarrow (4)$ is immediate.

We will complete the proof by showing that $(4)$ implies $(1)$. Note that the object $\operatorname{id}_{X} \in \operatorname{\mathcal{C}}_{X/}$ satisfies condition $(1)$ and therefore also satisfies condition $(3)$. It follows that there exists a commutative diagram

\[ \xymatrix@R =50pt@C=50pt{ \operatorname{\mathcal{C}}_{X/} \ar [rr]^-{F} \ar [dr] & & \operatorname{\mathcal{D}}\ar [dl]^{U} \\ & \operatorname{\mathcal{C}}& } \]

satisfying $F( \operatorname{id}_{X} ) = \widetilde{X}$. To complete the proof, it will suffice to show that if condition $(4)$ is satisfied, then $F$ is an equivalence of left fibrations over $\operatorname{\mathcal{C}}$. For every left fibration $V: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$, we have a commutative diagram of sets

\[ \xymatrix@R =50pt@C=50pt{ \pi _0( \operatorname{Fun}_{ / \operatorname{\mathcal{C}}}( \operatorname{\mathcal{D}}, \operatorname{\mathcal{E}}) ) \ar [rr]^-{\circ [F]} \ar [dr] & & \pi _0( \operatorname{Fun}_{ / \operatorname{\mathcal{C}}}( \operatorname{\mathcal{C}}_{X/}, \operatorname{\mathcal{E}}) ) \ar [dl] \\ & \pi _0( \{ X\} \times _{\operatorname{\mathcal{C}}} \operatorname{\mathcal{E}}) & , } \]

where the vertical maps are given by evaluation on the objects $\widetilde{X} \in \operatorname{\mathcal{D}}$ and $\operatorname{id}_{X} \in \operatorname{\mathcal{C}}_{X/}$, and are therefore bijective. It follows that the horizontal map is also bijective. $\square$