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Tag 016C

Chapter 14: Simplicial Methods > Section 14.3: Simplicial objects

Remark 14.3.3. By abuse of notation we sometimes write $d_i : U_n \to U_{n - 1}$ instead of $d^n_i$, and similarly for $s_i : U_n \to U_{n + 1}$. The relations among the morphisms $d^n_i$ and $s^n_i$ may be expressed as follows:

  1. If $i < j$, then $d_i \circ d_j = d_{j - 1} \circ d_i$.
  2. If $i < j$, then $d_i \circ s_j = s_{j - 1} \circ d_i$.
  3. We have $\text{id} = d_j \circ s_j = d_{j + 1} \circ s_j$.
  4. If $i > j + 1$, then $d_i \circ s_j = s_j \circ d_{i - 1}$.
  5. If $i \leq j$, then $s_i \circ s_j = s_{j + 1} \circ s_i$.

This means that whenever the compositions on both the left and the right are defined then the corresponding equality should hold.

    The code snippet corresponding to this tag is a part of the file simplicial.tex and is located in lines 241–256 (see updates for more information).

    \begin{remark}
    \label{remark-relations}
    By abuse of notation we sometimes write $d_i : U_n \to U_{n - 1}$
    instead of $d^n_i$, and similarly for $s_i : U_n \to U_{n + 1}$.
    The relations among the morphisms $d^n_i$ and $s^n_i$
    may be expressed as follows:
    \begin{enumerate}
    \item If $i < j$, then $d_i \circ d_j = d_{j - 1} \circ d_i$.
    \item If $i < j$, then $d_i \circ s_j = s_{j - 1} \circ d_i$.
    \item We have $\text{id} = d_j \circ s_j = d_{j + 1} \circ s_j$.
    \item If $i > j + 1$, then $d_i \circ s_j = s_j \circ d_{i - 1}$.
    \item If $i \leq j$, then $s_i \circ s_j = s_{j + 1} \circ s_i$.
    \end{enumerate}
    This means that whenever the compositions on both the left and the
    right are defined then the corresponding equality should hold.
    \end{remark}

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