Definition 14.20.1. Let \mathcal{C} be a category. Let U be a simplicial object of \mathcal{C}. An augmentation \epsilon : U \to X of U towards an object X of \mathcal{C} is a morphism from U into the constant simplicial object X.
14.20 Augmentations
Lemma 14.20.2. Let \mathcal{C} be a category. Let X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}). Let U be a simplicial object of \mathcal{C}. To give an augmentation of U towards X is the same as giving a morphism \epsilon _0 : U_0 \to X such that \epsilon _0 \circ d^1_0 = \epsilon _0 \circ d^1_1.
Proof. Given a morphism \epsilon : U \to X we certainly obtain an \epsilon _0 as in the lemma. Conversely, given \epsilon _0 as in the lemma, define \epsilon _ n : U_ n \to X by choosing any morphism \alpha : [0] \to [n] and taking \epsilon _ n = \epsilon _0 \circ U(\alpha ). Namely, if \beta : [0] \to [n] is another choice, then there exists a morphism \gamma : [1] \to [n] such that \alpha and \beta both factor as [0] \to [1] \to [n]. Hence the condition on \epsilon _0 shows that \epsilon _ n is well defined. Then it is easy to show that (\epsilon _ n) : U \to X is a morphism of simplicial objects. \square
Lemma 14.20.3. Let \mathcal{C} be a category with fibred products. Let f : Y\to X be a morphism of \mathcal{C}. Let U be the simplicial object of \mathcal{C} whose nth term is the (n + 1)fold fibred product Y \times _ X Y \times _ X \ldots \times _ X Y. See Example 14.3.5. For any simplicial object V of \mathcal{C} we have
\begin{align*} \mathop{\mathrm{Mor}}\nolimits _{\text{Simp}(\mathcal{C})}(V, U) & = \mathop{\mathrm{Mor}}\nolimits _{\text{Simp}_1(\mathcal{C})}(\text{sk}_1 V, \text{sk}_1 U) \\ & = \{ g_0 : V_0 \to Y \mid f \circ g_0 \circ d^1_0 = f \circ g_0 \circ d^1_1\} \end{align*}
In particular we have U = \text{cosk}_1 \text{sk}_1 U.
Proof. Suppose that g : \text{sk}_1V \to \text{sk}_1U is a morphism of 1-truncated simplicial objects. Then the diagram
\xymatrix{ V_1 \ar@<1ex>[r]^{d^1_0} \ar@<-1ex>[r]_{d^1_1} \ar[d]_{g_1} & V_0 \ar[d]^{g_0} \\ Y \times _ X Y \ar@<1ex>[r]^{pr_1} \ar@<-1ex>[r]_{pr_0} & Y \ar[r] & X }
is commutative, which proves that the relation shown in the lemma holds. We have to show that, conversely, given a morphism g_0 satisfying the relation f \circ g_0 \circ d^1_0 = f \circ g_0 \circ d^1_1 we get a unique morphism of simplicial objects g : V \to U. This is done as follows. For any n \geq 1 let g_{n, i} = g_0 \circ V([0] \to [n], 0 \mapsto i) : V_ n \to Y. The equality above implies that f \circ g_{n, i} = f \circ g_{n, i + 1} because of the commutative diagram
\xymatrix{ [0] \ar[rd]_{0 \mapsto 0} \ar[rrrrrd]^{0 \mapsto i} \\ & [1] \ar[rrrr]^{0 \mapsto i, 1\mapsto i + 1} & & & & [n] \\ [0] \ar[ru]^{0 \mapsto 1} \ar[rrrrru]_{0 \mapsto i + 1} }
Hence we get (g_{n, 0}, \ldots , g_{n, n}) : V_ n \to Y \times _ X\ldots \times _ X Y = U_ n. We leave it to the reader to see that this is a morphism of simplicial objects. The last assertion of the lemma is equivalent to the first equality in the displayed formula of the lemma. \square
Remark 14.20.4. Let \mathcal{C} be a category with fibre products. Let V be a simplicial object. Let \epsilon : V \to X be an augmentation. Let U be the simplicial object whose nth term is the (n + 1)st fibred product of V_0 over X. By a simple combination of Lemmas 14.20.2 and 14.20.3 we obtain a canonical morphism V \to U.
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