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 $n$th 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 In particular we have $U = \text{cosk}_1 \text{sk}_1 U$.
\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*}
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 $n$th 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 $V \to U$.
Post a comment
Your email address will not be published. Required fields are marked.
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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)