The Stacks project

Lemma 14.19.6. 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 there exists an $(n + 1)$-truncated simplicial object $\tilde U$, with $\text{sk}_ n \tilde U = U$ and $\tilde U_{n + 1} = U_{n + 1}$ such that the following adjointness holds

\[ \mathop{\mathrm{Mor}}\nolimits _{\text{Simp}_{n + 1}(\mathcal{C})}(V, \tilde U) = \mathop{\mathrm{Mor}}\nolimits _{\text{Simp}_ n(\mathcal{C})}(\text{sk}_ nV, U) \]

Proof. By Lemma 14.19.3 there are identifications

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

for $0 \leq i \leq n$. By Lemma 14.19.5 we have

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

Thus we may define for any $\varphi : [i] \to [j]$ with $i, j \leq n + 1$ the corresponding map $\tilde U(\varphi ) : \tilde U_ j \to \tilde U_ i$ exactly as in Lemma 14.19.2. This defines an $(n + 1)$-truncated simplicial object $\tilde U$ with $\text{sk}_ n \tilde U = U$.

To see the adjointness we argue as follows. Given any element $\gamma : \text{sk}_ n V \to U$ of the right hand side of the formula consider the morphisms $f_ i = \gamma _ n \circ d^{n + 1}_ i : V_{n + 1} \to V_ n \to U_ n$. These clearly satisfy the relations $d^ n_{j - 1} \circ f_ i = d^ n_ i \circ f_ j$ and hence define a unique morphism $V_{n + 1} \to U_{n + 1}$ by our choice of $U_{n + 1}$. Conversely, given a morphism $\gamma ' : V \to \tilde U$ of the left hand side we can simply restrict to $\Delta _{\leq n}$ to get an element of the right hand side. We leave it to the reader to show these are mutually inverse constructions. $\square$

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