Kerodon

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

Warning 10.2.6.27. Let $X_{\bullet }$ be a simplicial object of an $\infty $-category $\operatorname{\mathcal{C}}$. It follows from Remark 10.2.6.26 that every splitting of $X_{\bullet }$ determines a right homotopy inverse to the comparison map $T_{\bullet }: \operatorname{Dec}(X)_{\bullet } \rightarrow X_{\bullet }$. Beware that, in general, not every right homotopy inverse can be obtained in this way. For example, suppose that $\operatorname{\mathcal{C}}$ is (the nerve of) an ordinary category. Unwinding the definitions, we see that a morphism of simplicial objects from $X_{\bullet }$ to $\operatorname{Dec}(X)_{\bullet }$ is given by a collection of morphisms $h_{n}: X_{n} \rightarrow \operatorname{Dec}(X)_{n} = X_{n+1}$ which satisfy the identities

\[ d^{n+1}_{i} \circ h_ n = h_{n-1} \circ d^{n}_{i-1} \quad \quad s^{n+1}_{i} \circ h_{n} = h_{n+1} \circ s^{n}_{i-1} \]

for $0 < i \leq n+1$. Moreover, $h_{\bullet }$ is a right inverse of $T_{\bullet }$ if and only if it satisfies the further identity $d^{n+1}_{0} \circ h_ n = \operatorname{id}_{ X_ n }$ for each $n \geq 0$. However, $h_{\bullet }$ arises from a splitting of the simplicial object $X_{\bullet }$ only if it also satisfies the identities $s^{n+1}_{0} \circ h_{n} = h_{n+1} \circ h_{n}$; see Exercise 10.2.6.5 (compare with Warning 10.2.6.8).