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Corollary Let $q: X \rightarrow S$ be a cartesian fibration of simplicial sets, and let $X' \subseteq X$ be the simplicial subset spanned by those simplices $\sigma : \Delta ^ n \rightarrow X$ which carry each edge of $\Delta ^ n$ to a $q$-cartesian edge of $X$. Then the morphism $q|_{X'}: X' \rightarrow S$ is a right fibration of simplicial sets.

Proof. Choose integers $0 < i \leq n$; we wish to show that every lifting problem

\[ \xymatrix@R =50pt@C=50pt{ \Lambda ^{n}_{i} \ar [r]^-{ \sigma _0 } \ar [d] & X' \ar [d]^-{ q|_{X'} } \\ \Delta ^ n \ar@ {-->}[ur]^{\sigma } \ar [r]^-{\overline{\sigma } } & S } \]

admits a solution. In the special case $i = n = 1$, this follows immediately from our assumption that $q$ is a cartesian fibration. Assume therefore that $n \geq 2$. We first show that $\sigma _0$ can be extended to an $n$-simplex $\sigma : \Delta ^ n \rightarrow X$ satisfying $q \circ \sigma = \overline{\sigma }$. For $i < n$, this follows from the assumption that $q$ is an inner fibration. For $i = n$, it follows from the assumption that the edge

\[ \Delta ^1 \simeq \operatorname{N}_{\bullet }( \{ n-1 < n \} ) \hookrightarrow \Lambda ^ n_ i \xrightarrow { \sigma _0 } X' \hookrightarrow X \]

is $q$-cartesian. We now complete the proof by showing that $\sigma $ factors through the simplicial subset $X'$; that is, it carries each edge of $\Delta ^ n$ to a $q$-cartesian edge of $X$. For $n \geq 3$, this is immediate (since every edge of $\Delta ^ n$ is contained in $\Lambda ^{n}_{i}$). The case $n=2$ follows from Proposition $\square$