The Stacks project

Lemma 14.33.2. In Example 14.33.1 if

\[ 1_ Y = (d \star 1_ Y) \circ s = (1_ Y \star d) \circ s \quad \text{and}\quad (s \star 1) \circ s = (1 \star s) \circ s \]

then $X = (X_ n, d^ n_ j, s^ n_ j)$ is a simplicial object in the category of endofunctors of $\mathcal{C}$ and $d : X_0 = Y \to \text{id}_\mathcal {C}$ defines an augmentation.

Proof. To see that we obtain a simplicial object we have to check that the relations (1)(a) – (e) of Lemma 14.3.2 are satisfied. We will use the short hand notation

\[ 1_ a = 1_{X_{a - 1}} = 1_ Y \star \ldots \star 1_ Y \quad (a\text{ factors}) \]

for $a \geq 0$. With this notation we have

\[ d^ n_ j = 1_ j \star d \star 1_{n - j} \quad \text{and}\quad s^ n_ j = 1_ j \star s \star 1_{n - j} \]

We are repeatedly going to use the rule that for transformations of funtors $a, a', b, b'$ we have $(a' \circ a) \star (b' \circ b) = (a' \star b') \circ (a \star b)$ provided that the $\star $ and $\circ $ compositions in this formula make sense, see Categories, Lemma 4.28.2.

Condition (1)(a) always holds (no conditions needed on $d$ and $s$). Namely, let $0 \leq i < j \leq n + 1$. We have to show that $d^ n_ i \circ d^{n + 1}_ j = d^ n_{j - 1} \circ d^{n + 1}_ i$, i.e.,

\[ (1_ i \star d \star 1_{n - i}) \circ (1_ j \star d \star 1_{n + 1 - j}) = (1_{j - 1} \star d \star 1_{n + 1 - j}) \circ (1_ i \star d \star 1_{n + 1 - i}) \]

We can rewrite the left hand side as

\begin{align*} & (1_ i \star d \star 1_{j - i - 1} \star 1_{n + 1 - j}) \circ (1_ i \star 1_1 \star 1_{j - i - 1} \star d \star 1_{n + 1 - j}) \\ & = 1_ i \star \left((d \star 1_{j - i - 1}) \circ (1_1 \star 1_{j - i - 1} \star d)\right) \star 1_{n + 1 - j} \\ & = 1_ i \star d \star 1_{j - i - 1} \star d \star 1_{n + 1 - j} \end{align*}

The second equality is true because $d \circ 1_1 = d$ and $1_{j - i} \circ (1_{j - i - 1} \star d) = 1_{j - i - 1} \star d$. A similar computation gives the same result for the right hand side.

We check condition (1)(b). Let $0 \leq i < j \leq n - 1$. We have to show that $d^ n_ i \circ s^{n - 1}_ j = s^{n - 2}_{j - 1} \circ d^{n - 1}_ i$, i.e.,

\[ (1_ i \star d \star 1_{n - i}) \circ (1_ j \star s \star 1_{n - 1 - j}) = (1_{j - 1} \star s \star 1_{n - 1 - j}) \circ (1_ i \star d \star 1_{n - 1 - i}) \]

By the same kind of calculus as in case (1)(a) both sides simplify to $1_ i \star d \star 1_{j - i - 1} \star s \star 1_{n - j - 1}$.

We check condition (1)(c). Let $0 \leq j \leq n - 1$. We have to show $\text{id} = d^ n_ j \circ s^{n - 1}_ j = d^ n_{j + 1} \circ s^{n - 1}_ j$, i.e.,

\[ 1_ n = (1_ j \star d \star 1_{n - j}) \circ (1_ j \star s \star 1_{n - 1 - j}) = (1_{j + 1} \star d \star 1_{n - j - 1}) \circ (1_ j \star s \star 1_{n - 1 - j}) \]

This is easily seen to be implied by the first assumption of the lemma.

We check condition (1)(d). Let $0 < j + 1 < i \leq n$. We have to show $d^ n_ i \circ s^{n - 1}_ j = s^{n - 2}_ j \circ d^{n - 1}_{i - 1}$, i.e.,

\[ (1_ i \star d \star 1_{n - i}) \circ (1_ j \star s \star 1_{n - 1 - j}) = (1_ j \star s \star 1_{n - 2 - j}) \circ (1_{i - 1} \star d \star 1_{n - i}) \]

By the same kind of calculus as in case (1)(a) both sides simplify to $1_ j \star s \star 1_{i - j - 2} \star d \star 1_{n - i}$.

We check condition (1)(e). Let $0 \leq i \leq j \leq n - 1$. We have to show that $s^ n_ i \circ s^{n - 1}_ j = s^ n_{j + 1} \circ s^{n - 1}_ i$, i.e.,

\[ (1_ i \star s \star 1_{n - i}) \circ (1_ j \star s \star 1_{n - 1 - j}) = (1_{j + 1} \star s \star 1_{n - 1 - j}) \circ (1_ i \star s \star 1_{n - 1 - i}) \]

By the same kind of calculus as in case (1)(a) this reduces to

\[ (s \star 1_{j - i + 1}) \circ (1_{j - i} \star s) = (1_{j - i + 1} \star s) \circ (s \star 1_{j - i}) \]

If $j = i$ this is exactly one of the two assumptions of the lemma. For $j > i$ left and right hand side both reduce to the equality $s \star 1_{j - i - 1} \star s$ by calculations similar to those we did in case (1)(a).

Finally, in order to show that $d$ defines an augmentation we have to show that $d \circ (1_1 \star d) = d \circ (d \star 1_1)$ which is true because both sides are equal to $d \star d$. $\square$

Comments (0)

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0G5N. Beware of the difference between the letter 'O' and the digit '0'.



View Lemma 14.33.2 as pdf