Lemma 14.2.3. The morphisms $\delta ^ n_ j$ and $\sigma ^ n_ j$ satisfy the following relations.

1. If $0 \leq i < j \leq n + 1$, then $\delta ^{n + 1}_ j \circ \delta ^ n_ i = \delta ^{n + 1}_ i \circ \delta ^ n_{j - 1}$. In other words the diagram

$\xymatrix{ & [n] \ar[rd]^{\delta ^{n + 1}_ j} & \\ [n - 1] \ar[ru]^{\delta ^ n_ i} \ar[rd]_{\delta ^ n_{j - 1}} & & [n + 1] \\ & [n] \ar[ru]_{\delta ^{n + 1}_ i} & }$

commutes.

2. If $0 \leq i < j \leq n - 1$, then $\sigma ^{n - 1}_ j \circ \delta ^ n_ i = \delta ^{n - 1}_ i \circ \sigma ^{n - 2}_{j - 1}$. In other words the diagram

$\xymatrix{ & [n] \ar[rd]^{\sigma ^{n - 1}_ j} & \\ [n - 1] \ar[ru]^{\delta ^ n_ i} \ar[rd]_{\sigma ^{n - 2}_{j - 1}} & & [n - 1] \\ & [n - 2] \ar[ru]_{\delta ^{n - 1}_ i} & }$

commutes.

3. If $0 \leq j \leq n - 1$, then $\sigma ^{n - 1}_ j \circ \delta ^ n_ j = \text{id}_{[n - 1]}$ and $\sigma ^{n - 1}_ j \circ \delta ^ n_{j + 1} = \text{id}_{[n - 1]}$. In other words the diagram

$\xymatrix{ & [n] \ar[rd]^{\sigma ^{n - 1}_ j} & \\ [n - 1] \ar[ru]^{\delta ^ n_ j} \ar[rd]_{\delta ^ n_{j + 1}} \ar[rr]^{\text{id}_{[n - 1]}} & & [n - 1] \\ & [n] \ar[ru]_{\sigma ^{n - 1}_ j} & }$

commutes.

4. If $0 < j + 1 < i \leq n$, then $\sigma ^{n - 1}_ j \circ \delta ^ n_ i = \delta ^{n - 1}_{i - 1} \circ \sigma ^{n - 2}_ j$. In other words the diagram

$\xymatrix{ & [n] \ar[rd]^{\sigma ^{n - 1}_ j} & \\ [n - 1] \ar[ru]^{\delta ^ n_ i} \ar[rd]_{\sigma ^{n - 2}_ j} & & [n - 1] \\ & [n - 2] \ar[ru]_{\delta ^{n - 1}_{i - 1}} & }$

commutes.

5. If $0 \leq i \leq j \leq n - 1$, then $\sigma ^{n - 1}_ j \circ \sigma ^ n_ i = \sigma ^{n - 1}_ i \circ \sigma ^ n_{j + 1}$. In other words the diagram

$\xymatrix{ & [n] \ar[rd]^{\sigma ^{n - 1}_ j} & \\ [n + 1] \ar[ru]^{\sigma ^ n_ i} \ar[rd]_{\sigma ^ n_{j + 1}} & & [n - 1] \\ & [n] \ar[ru]_{\sigma ^{n - 1}_ i} & }$

commutes.

Proof. Omitted. $\square$

There are also:

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