Lemma 14.33.4. In Situation 14.33.1 the system $X = (X_ n, d^ n_ j, s^ n_ j)$ is a simplicial object of $\text{Fun}(\mathcal{A}, \mathcal{A})$ and $\epsilon _0$ defines an augmentation $\epsilon$ from $X$ to the constant simplicial object with value $X_{-1} = \text{id}_\mathcal {A}$.

Proof. Suppose that we have shown that $X$ is a simplicial object. Then to prove that $\epsilon _0 = k$ defines an augmentation we have to check that $\epsilon _0 \circ d^1_0 = \epsilon _0 \circ d^1_1$ as morphisms $X_1 \to X_{-1}$, see Lemma 14.20.2. In other words, we have to check that the diagram

$\xymatrix{ F \circ i \circ F \circ i \ar[rr]_{1_{F \circ i} \star k} \ar[d]_{k \star 1_{F \circ i}} & & F \circ i \ar[d]^ k \\ F \circ i \ar[rr]^ k & & \text{id}_\mathcal {A} }$

is commutative. More precisely we should write this as the equality

$(k \star 1_{\text{id}_\mathcal {A}}) \circ (1_{F \circ i} \star k) = (1_{\text{id}_\mathcal {A}} \star k) \circ (k \star 1_{F \circ i})$

as morphisms $(F \circ i) \circ (F \circ i) \to \text{id}_\mathcal {A} \circ \text{id}_\mathcal {A}$. Applying the general property of Categories, Lemma 4.28.2 both sides expand to $k \star k$ when equality holds.

To prove that $X$ is a simplicial object we have to check (see Remark 14.3.3):

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 $s_ i \circ s_ j = s_{j + 1} \circ s_ i$.

Relation (1) is proved in exactly the same manner as the proof of the equality $\epsilon _0 \circ d^1_0 = \epsilon _0 \circ d^1_1$ above.

The simplest case of equality (5) is the commutativity of the diagram

$\xymatrix{ F \circ i \ar[rrr]_{1_ F \star t \star 1_ i} \ar[d]_{1_ F \star t \star 1_ i} & & & F \circ i \circ F \circ i \ar[d]^{1_ F \star t \star 1_{i \circ F \circ i}} \\ F \circ i \circ F \circ i \ar[rrr]^{1_{F \circ i \circ F} \star t \star 1_ i} & & & F \circ i \circ F \circ i \circ F \circ i }$

which holds because both compositions expand to the morphism $1_ F \star t \star t \star 1_ i$ from $F \circ \text{id}_\mathcal {A} \circ \text{id}_\mathcal {A} \circ i$ to $F \circ (i \circ F) \circ (i \circ F) \circ i$. All other cases of (5) are proved in the same manner.

The simplest case of equalities (2) and (4) is the commutativity of the diagram

$\xymatrix{ F \circ i \circ F \circ i \ar[rrr]_{1_ F \star k \star 1_ i} \ar[d]_{1_{F \circ i \circ F \circ i} \star t} & & & F \circ i \ar[d]_{1_{F \circ i} \star t} \\ F \circ i \circ F \circ i \circ F \circ i \ar[rrr]^{1_ F \star k \star 1_{i \circ F \circ i \circ F}} & & & F \circ i \circ F \circ i }$

which again holds because both compositions expand to give $1_ F \star k \star 1_ i \star t$ as maps from $F \circ (i \circ F) \circ i \circ \text{id}_\mathcal {A}$ to $F \circ \text{id}_\mathcal {A} \circ i \circ (F \circ i)$. All other cases of (2) and (4) are proved in the same manner.

The relations (3) are the only nontrivial ones and these are consequences of the fact that the compositions in (14.33.2.1) are the identity. For example, the simplest case of (3) states that the compositions

$\xymatrix{ F \circ i \ar[rr]_{1_{F \circ i} \star t} \ar[d]_{t \star 1_{F \circ i}} & & F \circ i \circ F \circ i \ar[d]^{1_ F \star k \star 1_ i} \\ F \circ i \circ F \circ i \ar[rr]^{1_ F \star k \star 1_ i} & & F \circ i }$

go around the diagram either way evaluate out to the identity. Going around the top the composition evaluates to $1_ F \star ((k \star 1_ i) \circ (1_ i \star t))$ which is the identity by what was said above. The other cases of (3) are proved in the same manner. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).