Proof.
Without loss of generality, we may assume that $f$ carries each edge of $K$ to a $U$-cocartesian morphism of $\operatorname{\mathcal{C}}$. Let $\pi : \Delta ^1 \times K^{\triangleright } \rightarrow K^{\triangleright }$ be the morphism which is the identity on $\{ 0\} \times K^{\triangleright }$ and which carries $\{ 1\} \times K^{\triangleright }$ to the cone point $v \in K^{\triangleright }$. Set $C = f(v) \in \operatorname{\mathcal{C}}$ and $D = U(C) \in \operatorname{\mathcal{D}}$. Proposition 5.2.1.3 guarantees that the lifting problem
\[ \xymatrix@R =50pt@C=50pt{ \{ 0\} \times K^{\triangleright } \ar [r]^-{ \overline{f}} \ar [d] & \operatorname{\mathcal{C}}\ar [d]^{U} \\ \Delta ^1 \times K^{\triangleright } \ar [r]^-{U \circ \overline{f} \circ \pi } \ar@ {-->}[ur]^{ \alpha } & \operatorname{\mathcal{D}}} \]
admits a solution $\alpha : \Delta ^1 \times K^{\triangleright } \rightarrow \operatorname{\mathcal{C}}$ which carries $\Delta ^1 \times \{ x\} $ to a $U$-cocartesian morphism of $\operatorname{\mathcal{C}}$, for each vertex $x \in K^{\triangleright }$. Set $\overline{g} = \alpha |_{ \{ 1\} \times K^{\triangleright }}$, which we regard as a morphism from $K^{\triangleright }$ to the $\infty $-category $\operatorname{\mathcal{C}}_{D}$, and let us identify $\alpha $ with a natural transformation from $\overline{f}$ to $\overline{g}$. Note that $\alpha _{v}: \overline{f}(v) \rightarrow \overline{g}(v)$ is a $U$-cocartesian morphism of $\operatorname{\mathcal{C}}$ satisfying $U(\alpha _ v) = \operatorname{id}_{D}$, and is therefore an isomorphism (Proposition 5.1.1.9). Applying Proposition 7.3.9.1, we can reformulate $(2)$ as follows:
- $(2')$
The morphism $\overline{g}: K^{\triangleright } \rightarrow \operatorname{\mathcal{C}}_{D}$ is a $U$-colimit diagram in $\operatorname{\mathcal{C}}$.
Set $g = \overline{g}|_{K}$. For every edge $u: x \rightarrow y$ of $K$, we have a commutative diagram
\[ \xymatrix@R =50pt@C=50pt{ f(x) \ar [r]^-{ f(u) } \ar [d]^{ \alpha _ x} & f(y) \ar [d]^{ \alpha _ y} \\ g(x) \ar [r]^-{ g(u) } & g(y) } \]
where $f(u)$, $\alpha _ x$, and $\alpha _ y$ are $U$-cocartesian. Applying Corollary 5.1.2.4, we deduce that $g(u)$ is $U$-cocartesian when viewed as a morphism of $\operatorname{\mathcal{C}}$, and is therefore an isomorphism in the $\infty $-category $\operatorname{\mathcal{C}}_{D}$ (Proposition 5.1.1.9). Similarly, for every vertex $x \in K$, the unique edge $c_{x}: x \rightarrow v$ of $K^{\triangleright }$ determines a commutative diagram
\[ \xymatrix@R =50pt@C=50pt{ f(x) \ar [r]^-{\overline{f}(c_ x)} \ar [d]^{ \alpha _ x} & \overline{f}(v) \ar [d]^{ \alpha _ v} \\ g(x) \ar [r]^-{ \overline{g}(c_ x)} & \overline{g}(v), } \]
where $\alpha _{x}$ is $U$-cocartesian and $\alpha _{v}$ is an isomorphism. Combining Corollary 5.1.2.4, Corollary 5.1.2.5, and Proposition 5.1.1.9, we see that $\overline{f}(c_ x)$ is $U$-cocartesian if and only if $\overline{g}(c_ x)$ is an isomorphism in the $\infty $-category $\operatorname{\mathcal{C}}_{D}$. We can therefore reformulate condition $(1)$ as follows:
- $(1')$
The diagram $\overline{g}$ carries each edge of $K^{\triangleright }$ to an isomorphism in the $\infty $-category $\operatorname{\mathcal{C}}_{D}$.
By virtue of Corollary 7.2.3.5, $(1')$ is equivalent to the requirement that $\overline{g}$ is a colimit diagram in the $\infty $-category $\operatorname{\mathcal{C}}_{D}$. In particular, the implication $(2') \Rightarrow (1')$ follows from Corollary 7.1.6.22. To prove the converse, it will suffice to show that condition $(1')$ is satisfied, then for every morphism $e: D \rightarrow D'$ in $\operatorname{\mathcal{D}}$, the covariant transport functor $e_{!}: \operatorname{\mathcal{C}}_{D} \rightarrow \operatorname{\mathcal{C}}_{D'}$ carries $\overline{g}$ to a colimit diagram in the $\infty $-category $\operatorname{\mathcal{C}}_{D'}$ (Proposition 7.3.9.2). This follows immediately from Corollary 7.2.3.5 (applied to the composite diagram $K^{\triangleleft } \xrightarrow { \overline{g} } \operatorname{\mathcal{C}}_{D} \xrightarrow { e_{!} } \operatorname{\mathcal{C}}_{D'}$).
$\square$