$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$

Proposition 7.1.1.11. Let $\operatorname{\mathcal{C}}$ be an $\infty $-category and let $u: K \rightarrow \operatorname{\mathcal{C}}$ be a diagram. Then:

Suppose that the diagram $u$ has limit $Y \in \operatorname{\mathcal{C}}$. Then an object $X \in \operatorname{\mathcal{C}}$ is a limit of $u$ if and only if it is isomorphic to $Y$.

Suppose that the diagram $u$ has colimit $Y \in \operatorname{\mathcal{C}}$. Then an object $X \in \operatorname{\mathcal{C}}$ is a colimit of $u$ if and only if it is isomorphic to $Y$.

**Proof.**
Let $\beta : \underline{Y} \rightarrow u$ be a natural transformation which exhibits $Y$ as a limit of the diagram $u$. For any object $X$ and any natural transformation $\alpha : \underline{X} \rightarrow u$, there exists a morphism $f: X \rightarrow Y$ such that $\alpha $ is a composition of $\beta $ with the constant natural transformation $\underline{f}: \underline{X} \rightarrow \underline{Y}$. If $\alpha $ also exhibits $X$ as a limit of the diagram $u$, then $f$ is an isomorphism (Remark 7.1.1.8); in particular, $X$ is isomorphic to $Y$. Conversely, if $f: X \rightarrow Y$ is an isomorphism, then any composition of $\underline{f}$ with $\beta $ is a natural transformation $\underline{X} \rightarrow u$ which exhibits $X$ as a limit of $u$ (Remark 7.1.1.8), so that $X$ is a limit of $Y$. This proves the first assertion; the proof of the second follows by applying the same argument to the opposite $\infty $-category $\operatorname{\mathcal{C}}^{\operatorname{op}}$.
$\square$