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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