# The Stacks Project

## Tag 016C

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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