Proposition 3.3.1.5. Let $X_{\bullet }$ and $Y_{\bullet }$ be simplicial sets, and suppose that $X_{\bullet }$ is braced. Then the restriction map
is a bijection.
Proof. Fix a morphism of semisimplicial sets $f_0: X_{\bullet }^{\mathrm{nd} } \rightarrow Y_{\bullet }$; we wish to show that $f_0$ extends uniquely to a morphism of simplicial sets from $X_{\bullet }$ to $Y_{\bullet }$. Let $\sigma $ be an $n$-simplex of $X_{\bullet }$. By virtue of Proposition 1.1.3.8, we can write $\sigma $ uniquely as $\alpha ^{\ast }(\tau )$, where $\alpha : [n] \twoheadrightarrow [m]$ is a nondecreasing surjection and $\tau $ is a nondegenerate $m$-simplex of $X_{\bullet }$. Define $f(\sigma ) = \alpha ^{\ast } f_0(\tau ) \in Y_{n}$. It is clear that any extension of $f_0$ to a morphism of simplicial sets $X_{\bullet } \rightarrow Y_{\bullet }$ must be given by the construction $\sigma \mapsto f(\sigma )$. It will therefore suffice to show that the construction $\sigma \mapsto f(\sigma )$ is a morphism of simplicial sets.
Let $\sigma $, $\tau $, and $\alpha $ be as above, and fix a nondecreasing map $\beta : [n'] \rightarrow [n]$. We wish to prove that $f( \beta ^{\ast } \sigma ) = \beta ^{\ast } f(\sigma )$ in the set $Y_{n'}$. Note that $(\alpha \circ \beta ): [n'] \rightarrow [m]$ factors uniquely as a composition $[n'] \xrightarrow { \alpha ' } [m'] \xrightarrow {\beta '} [m]$, where $\alpha '$ is surjective and $\beta '$ is injective. Since $X_{\bullet }$ is braced, $\beta '^{\ast }(\tau )$ is a nondegenerate $m'$-simplex of $X_{\bullet }$. We now compute
where the second and fifth equality follow from the identity $\alpha \circ \beta = \beta ' \circ \alpha '$, the third and sixth equality follow from the definition of $f$, and the fourth equality from the fact that $f_0$ is a morphism of semisimplicial sets. $\square$