Lemma 5.2.3.12. Let $U: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{E}}'$ be an inner fibration of simplicial sets. Then the induced map
is also an inner fibration of simplicial sets.
Lemma 5.2.3.12. Let $U: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{E}}'$ be an inner fibration of simplicial sets. Then the induced map
is also an inner fibration of simplicial sets.
Proof. Suppose we are given integers $0 < i < n$; we wish to show that every lifting problem
admits a solution. Let $\alpha $ denote the composite map
If $\alpha $ is a constant morphism, then the existence of $\sigma $ is immediate. We may therefore assume without loss of generality that $\alpha $ is not constant. Write $\sigma _0 = (\alpha _0, \tau _0)$, where $\alpha _0 = \alpha |_{ \Lambda ^{n}_{i} }$ and $\tau _0: \Lambda ^{n}_{i} \rightarrow \operatorname{\mathcal{E}}$ is a morphism of simplicial sets, and let $\overline{\tau }$ denote the composite map $\Delta ^ n \xrightarrow { \overline{\sigma } } \operatorname{\mathcal{E}}\star _{\operatorname{\mathcal{E}}'} \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{E}}'$. Since $U$ is an inner fibration, the lifting problem
admits a solution. We now observe that the pair $\sigma = (\alpha , \tau )$ can be regarded as an $n$-simplex of $\Delta ^1 \times \operatorname{\mathcal{E}}$ which solves the lifting problem (5.18). $\square$