Remark 14.19.7. Let $U$, and $U_{n + 1}$ be as in Lemma 14.19.6. On $T$-valued points we can easily describe the face and degeneracy maps of $\tilde U$. Explicitly, the maps $d^{n + 1}_ i : U_{n + 1} \to U_ n$ are given by

$(f_0, \ldots , f_{n + 1}) \longmapsto f_ i.$

And the maps $s^ n_ j : U_ n \to U_{n + 1}$ are given by

\begin{eqnarray*} f & \longmapsto & ( s^{n - 1}_{j - 1} \circ d^{n - 1}_0 \circ f, \\ & & s^{n - 1}_{j - 1} \circ d^{n - 1}_1 \circ f, \\ & & \ldots \\ & & s^{n - 1}_{j - 1} \circ d^{n - 1}_{j - 1} \circ f, \\ & & f, \\ & & f, \\ & & s^{n - 1}_ j \circ d^{n - 1}_{j + 1} \circ f, \\ & & s^{n - 1}_ j \circ d^{n - 1}_{j + 2} \circ f, \\ & & \ldots \\ & & s^{n - 1}_ j \circ d^{n - 1}_ n \circ f ) \end{eqnarray*}

where we leave it to the reader to verify that the RHS is an element of the displayed set of Lemma 14.19.6. For $n = 0$ there is one map, namely $f \mapsto (f, f)$. For $n = 1$ there are two maps, namely $f \mapsto (f, f, s_0d_1f)$ and $f \mapsto (s_0d_0f, f, f)$. For $n = 2$ there are three maps, namely $f \mapsto (f, f, s_0d_1f, s_0d_2f)$, $f \mapsto (s_0d_0f, f, f, s_1d_2f)$, and $f \mapsto (s_1d_0f, s_1d_1f, f, f)$. And so on and so forth.

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