Lemma 14.19.5. Let $n$ be an integer $\geq 1$. Let $U$ be a $n$-truncated simplicial object of $\mathcal{C}$. Consider the contravariant functor from $\mathcal{C}$ to $\textit{Sets}$ which associates to an object $T$ the set

\[ \{ (f_0, \ldots , f_{n + 1}) \in \mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(T, U_ n) \mid d^ n_{j - 1} \circ f_ i = d^ n_ i \circ f_ j \ \forall \ 0\leq i < j\leq n + 1\} \]

If this functor is representable by some object $U_{n + 1}$ of $\mathcal{C}$, then

\[ U_{n + 1} = \mathop{\mathrm{lim}}\nolimits _{(\Delta /[n + 1])_{\leq n}^{opp}} U(n) \]

**Proof.**
The limit, if it exists, represents the functor that associates to an object $T$ the set

\[ \{ (f_\alpha )_{\alpha : [k] \to [n + 1], k \leq n} \mid f_{\alpha \circ \psi } = U(\psi ) \circ f_\alpha \ \forall \ \psi : [k'] \to [k], \alpha : [k] \to [n + 1] \} . \]

In fact we will show this functor is isomorphic to the one displayed in the lemma. The map in one direction is given by the rule

\[ (f_\alpha )_{\alpha } \longmapsto (f_{\delta ^{n + 1}_0}, \ldots , f_{\delta ^{n + 1}_{n + 1}}). \]

This satisfies the conditions of the lemma because

\[ d^ n_{j - 1} \circ f_{\delta ^{n + 1}_ i} = f_{\delta ^{n + 1}_ i \circ \delta ^ n_{j - 1}} = f_{\delta ^{n + 1}_ j \circ \delta ^ n_ i} = d^ n_ i \circ f_{\delta ^{n + 1}_ j} \]

by the relations we recalled above the lemma. To construct a map in the other direction we have to associate to a system $(f_0, \ldots , f_{n + 1})$ as in the displayed formula of the lemma a system of maps $f_\alpha $. Let $\alpha : [k] \to [n + 1]$ be given. Since $k \leq n$ the map $\alpha $ is not surjective. Hence we can write $\alpha = \delta ^{n + 1}_ i \circ \psi $ for some $0 \leq i \leq n + 1$ and some $\psi : [k] \to [n]$. We have no choice but to define

\[ f_\alpha = U(\psi ) \circ f_ i. \]

Of course we have to check that this is independent of the choice of the pair $(i, \psi )$. First, observe that given $i$ there is a unique $\psi $ which works. Second, suppose that $(j, \phi )$ is another pair. Then $i \not= j$ and we may assume $i < j$. Since both $i, j$ are not in the image of $\alpha $ we may actually write $\alpha = \delta ^{n + 1}_{i, j} \circ \xi $ and then we see that $\psi = \delta ^ n_{j - 1} \circ \xi $ and $\phi = \delta ^ n_ i \circ \xi $. Thus

\begin{eqnarray*} U(\psi ) \circ f_ i & = & U(\delta ^ n_{j - 1} \circ \xi ) \circ f_ i \\ & = & U(\xi ) \circ d^ n_{j - 1} \circ f_ i \\ & = & U(\xi ) \circ d^ n_ i \circ f_ j \\ & = & U(\delta ^ n_ i \circ \xi ) \circ f_ j \\ & = & U(\phi ) \circ f_ j \end{eqnarray*}

as desired. We still have to verify that the maps $f_\alpha $ so defined satisfy the rules of a system of maps $(f_\alpha )_\alpha $. To see this suppose that $\psi : [k'] \to [k]$, $\alpha : [k] \to [n + 1]$ with $k, k' \leq n$. Set $\alpha ' = \alpha \circ \psi $. Choose $i$ not in the image of $\alpha $. Then clearly $i$ is not in the image of $\alpha '$ also. Write $\alpha = \delta ^{n + 1}_ i \circ \phi $ (we cannot use the letter $\psi $ here because we've already used it). Then obviously $\alpha ' = \delta ^{n + 1}_ i \circ \phi \circ \psi $. By construction above we then have

\[ U(\psi ) \circ f_\alpha = U(\psi ) \circ U(\phi ) \circ f_ i = U(\phi \circ \psi ) \circ f_ i = f_{\alpha \circ \psi } = f_{\alpha '} \]

as desired. We leave to the reader the pleasant task of verifying that our constructions are mutually inverse bijections, and are functorial in $T$.
$\square$

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