Proposition 10.2.2.11. Let $\operatorname{\mathcal{C}}$ be a category, let $Y$ be an object of $\operatorname{\mathcal{C}}$, and let $\underline{Y}$ denote the constant semisimplicial object of $\operatorname{\mathcal{C}}$ taking the value $Y$. For every semisimplicial object $X_{\bullet }$ of $\operatorname{\mathcal{C}}$, the evaluation map
\[ \operatorname{Hom}_{ \operatorname{Fun}( \operatorname{{\bf \Delta }}^{\operatorname{op}}_{\operatorname{inj}}, \operatorname{\mathcal{C}}) }( X_{\bullet }, \underline{Y} ) \rightarrow \operatorname{Hom}_{\operatorname{\mathcal{C}}}( X_0, Y ) \]
is a monomorphism, whose image is the set of morphisms $\epsilon : X_0 \rightarrow Y$ which satisfy the following condition:
- $(\ast )$
The face operators $d^{1}_0, d^{1}_1: X_1 \rightarrow X_0$ of the simplicial object $X_{\bullet }$ satisfy $\epsilon \circ d^{1}_0 = \epsilon \circ d^{1}_1$.
Proof.
For every integer $n \geq 0$, let $\iota _{n}$ denote the inclusion map $[0] = \{ 0 \} \hookrightarrow \{ 0 < 1 < \cdots < n \} = [n]$ and write $\iota _{n}^{\ast }: X_{n} \rightarrow X_0$ for the associated morphism of $\operatorname{\mathcal{C}}$. If $f_{\bullet }: X_{\bullet } \rightarrow \underline{Y}$ is a morphism of semisimplicial objects, then we must have $f_{n} = f_0 \circ \iota ^{\ast }_{n}$ for each $n \geq 0$; in particular, $f_{\bullet }$ is uniquely determined by the morphism $\epsilon = f_0$. To complete the proof, it will suffice to show that if a morphism $\epsilon : X_0 \rightarrow Y$ satisfies condition $(\ast )$, then the collection $\{ (\epsilon \circ \iota ^{\ast }_ n): X_ n \rightarrow Y \} _{n \geq 0}$ determines a morphism of semisimplicial objects from $X_{\bullet }$ to $\underline{Y}$ (the converse follows immediately from the definitions). Fix a strictly increasing function $\alpha : [m] \hookrightarrow [n]$; we wish to show that the diagram
10.3
\begin{equation} \begin{gathered}\label{equation:maps-into-constant-simplicial-easy} \xymatrix@R =50pt@C=50pt{ X_{n} \ar [d]^{ \alpha ^{\ast } } \ar [r]^-{ \iota ^{\ast }_{n} } & X_{0} \ar [d]^{ \epsilon } \\ X_{m} \ar [r]^-{ \epsilon \circ \iota ^{\ast }_{m} } & Y } \end{gathered} \end{equation}
commutes. If $\alpha (0) = 0$, then $\iota _{n} = \alpha \circ \iota _{m}$. It follows that $\iota _{n}^{\ast } = \iota _{m}^{\ast } \circ \alpha ^{\ast }$, and the desired result follows by composing with $\epsilon $ on both sides. We may therefore assume without loss of generality that $\alpha (0) > 0$. Let $\beta : [1] \hookrightarrow [n]$ be the strictly increasing function given by $\beta (0) = 0$ and $\beta (1) = \alpha (0)$. Then (10.3) can be identified with the outer rectangle of the diagram
\[ \xymatrix@R =50pt@C=50pt{ X_{n} \ar [d]^{ \alpha ^{\ast } } \ar [r]^-{ \beta ^{\ast } } & X_{1} \ar [r]^-{ d^{1}_1 } \ar [d]^{d^{1}_0} & X_0 \ar [d]^{ \epsilon } \\ X_{m} \ar [r]^-{ \iota ^{\ast }_ m } & X_0 \ar [r]^-{\epsilon } & Y, } \]
where the left square commutes by the naturality of the construction $[k] \mapsto X_{k}$, and the right square commutes by virtue of assumption $(\ast )$.
$\square$