Lemma 14.19.15. The canonical map $\Delta [n] \to \text{cosk}_1 \text{sk}_1 \Delta [n]$ is an isomorphism.

Proof. Consider a simplicial set $U$ and a morphism $f : U \to \Delta [n]$. This is a rule that associates to each $u \in U_ i$ a map $f_ u : [i] \to [n]$ in $\Delta$. Furthermore, these maps should have the property that $f_ u \circ \varphi = f_{U(\varphi )(u)}$ for any $\varphi : [j] \to [i]$. Denote $\epsilon ^ i_ j : [0] \to [i]$ the map which maps $0$ to $j$. Denote $F : U_0 \to [n]$ the map $u \mapsto f_ u(0)$. Then we see that

$f_ u(j) = F(\epsilon ^ i_ j(u))$

for all $0 \leq j \leq i$ and $u \in U_ i$. In particular, if we know the function $F$ then we know the maps $f_ u$ for all $u\in U_ i$ all $i$. Conversely, given a map $F : U_0 \to [n]$, we can set for any $i$, and any $u \in U_ i$ and any $0 \leq j \leq i$

$f_ u(j) = F(\epsilon ^ i_ j(u))$

This does not in general define a morphism $f$ of simplicial sets as above. Namely, the condition is that all the maps $f_ u$ are nondecreasing. This clearly is equivalent to the condition that $F(\epsilon ^ i_ j(u)) \leq F(\epsilon ^ i_{j'}(u))$ whenever $0 \leq j \leq j' \leq i$ and $u \in U_ i$. But in this case the morphisms

$\epsilon ^ i_ j, \epsilon ^ i_{j'} : [0] \to [i]$

both factor through the map $\epsilon ^ i_{j, j'} : [1] \to [i]$ defined by the rules $0 \mapsto j$, $1 \mapsto j'$. In other words, it is enough to check the inequalities for $i = 1$ and $u \in X_1$. In other words, we have

$\mathop{Mor}\nolimits (U, \Delta [n]) = \mathop{Mor}\nolimits (\text{sk}_1 U, \text{sk}_1 \Delta [n])$

as desired. $\square$

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