Kerodon

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

Proposition 4.3.3.24. Let $u: X \rightarrow X'$ and $v: Y \rightarrow Y'$ be inner fibrations of simplicial sets. Then the join $(u \star v): X \star Y \rightarrow X' \star Y'$ is also an inner fibration of simplicial sets.

Proof of Proposition 4.3.3.24. Let $u: X \rightarrow X'$ and $v: Y \rightarrow Y'$ be inner fibrations of simplicial sets and let $0 < i < n$ be integers; we wish to show that every lifting problem

4.14
\begin{equation} \begin{gathered}\label{equation:join-of-inner-fibration} \xymatrix@R =50pt@C=50pt{ \Lambda ^{n}_{i} \ar [r]^-{ \sigma _0 } \ar [d] & X \star Y \ar [d]^{u \star v } \\ \Delta ^{n} \ar [r]^-{ \sigma ' } \ar@ {-->}[ur]^{ \sigma } & X' \star Y' } \end{gathered} \end{equation}

admits a solution. If $\sigma '$ factors through either $X'$ or $Y'$, this follows immediately from our assumption that $u$ and $v$ are inner fibrations. We may therefore assume without loss of generality that $\sigma '$ factors as a composition

\[ \Delta ^{n} = \Delta ^{p+1+q} \simeq \Delta ^{p} \star \Delta ^{q} \xrightarrow { \sigma '_{-} \star \sigma '_{+} } X' \star Y' \]

for some pair of integers $p,q \geq 0$ satisfying $p+1+q=n$ and simplices $\sigma '_{-}: \Delta ^{p} \rightarrow X'$ and $\sigma '_{+}: \Delta ^{q} \rightarrow Y'$. Let $\iota _{-}$ denote the inclusion map

\[ \Delta ^{p} \hookrightarrow \Delta ^{p} \star \Delta ^{q} \simeq \Delta ^{p+1+q} = \Delta ^{n}, \]

and define $\iota _{+}: \Delta ^{q} \hookrightarrow \Delta ^ n$ similarly. Note that both $\iota _{-}$ and $\iota _{+}$ factor through the inner horn $\Lambda ^{n}_{i} \subseteq \Delta ^ n$. Set $\sigma _{-} = \sigma _0 \circ \iota _{-}$ and $\sigma _{+} = \sigma _0 \circ \iota _{+}$. Unwinding the definitions, we see that the composite map

\[ \Delta ^{n} = \Delta ^{p+1+q} \simeq \Delta ^{p} \star \Delta ^{q} \xrightarrow { \sigma _{-} \star \sigma _{+} } X \star Y \]

determines an $n$-simplex $\sigma $ of $X \star Y$ which is a solution to the lifting problem (4.14). $\square$