The Stacks project

Lemma 14.23.3. Let $\mathcal{A}$ be an abelian category. Let $A$ be an object of $\mathcal{A}$ and let $k$ be an integer. We have $H_ i(s(K(A, k))) = A$ if $i = k$ and $0$ else.

Proof. First, let us prove this if $k = 0$. In this case we have $K(A, 0)_ n = A$ for all $n$. Furthermore, all the maps in this simplicial abelian group are $\text{id}_ A$, in other words $K(A, 0)$ is the constant simplicial object with value $A$. The boundary maps $d_ n = \sum _{i = 0}^ n (-1)^ i \text{id}_ A = 0$ if $n$ odd and $ = \text{id}_ A$ if $n$ is even. Thus $s(K(A, 0))$ looks like this

\[ \ldots \to A \xrightarrow {0} A \xrightarrow {1} A \xrightarrow {0} A \to 0 \]

and the result is clear.

Next, we prove the result for all $k$ by induction. Given the result for $k$ consider the short exact sequence

\[ 0 \to K(A, k) \to E \to K(A, k + 1) \to 0 \]

from Lemma 14.22.4. By Lemma 14.22.1 the associated sequence of chain complexes is exact. By Lemma 14.23.2 we see that $s(E)$ is acyclic. Hence the result for $k + 1$ follows from the long exact sequence of homology, see Homology, Lemma 12.13.6. $\square$


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Your email address will not be published.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 0197. Beware of the difference between the letter 'O' and the digit '0'.