Lemma 14.26.2. In the situation above, we have the following relations:

1. We have $h_{n, 0} = b_ n$ and $h_{n, n + 1} = a_ n$.

2. We have $d^ n_ j \circ h_{n, i} = h_{n - 1, i - 1} \circ d^ n_ j$ for $i > j$.

3. We have $d^ n_ j \circ h_{n, i} = h_{n - 1, i} \circ d^ n_ j$ for $i \leq j$.

4. We have $s^ n_ j \circ h_{n, i} = h_{n + 1, i + 1} \circ s^ n_ j$ for $i > j$.

5. We have $s^ n_ j \circ h_{n, i} = h_{n + 1, i} \circ s^ n_ j$ for $i \leq j$.

Conversely, given a system of maps $h_{n, i}$ satisfying the properties listed above, then these define a morphism $h$ which is a homotopy from $a$ to $b$.

Proof. Omitted. You can prove the last statement using the fact, see Lemma 14.2.4 that to give a morphism of simplicial objects is the same as giving a sequence of morphisms $h_ n$ commuting with all $d^ n_ j$ and $s^ n_ j$. $\square$

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