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Proposition Let $\operatorname{\mathcal{C}}$ be an $\infty $-category. Then the opposite simplicial set $\operatorname{\mathcal{C}}^{\operatorname{op}}$ is also an $\infty $-category.

Proof. Let $\sigma _0: \Lambda ^{n}_{i} \rightarrow \operatorname{\mathcal{C}}^{\operatorname{op}}$ be a map of simplicial sets for $0 < i < n$; we wish to show that $\sigma _0$ can be extended to an $n$-simplex of $\operatorname{\mathcal{C}}^{\operatorname{op}}$. Passing to opposite simplicial sets, we are reduced to showing that the map $\sigma _0^{\operatorname{op}}: ( \Lambda ^{n}_{i})^{\operatorname{op}} \rightarrow \operatorname{\mathcal{C}}$ can be extended to a map $( \Delta ^ n )^{\operatorname{op}} \rightarrow \operatorname{\mathcal{C}}$. This follows from our assumption that $\operatorname{\mathcal{C}}$ is an $\infty $-category, since there is a unique isomorphism $( \Delta ^ n)^{\operatorname{op}} \simeq \Delta ^ n$ which carries the simplicial subset $(\Lambda ^{n}_{i})^{\operatorname{op}}$ to $\Lambda ^{n}_{n-i}$. $\square$