Lemma 14.23.4. Let $\mathcal{A}$ be an abelian category. Let $U$ be a simplicial object of $\mathcal{A}$. The canonical map $N(U_ n) \to U_ n$ gives rise to a morphism of complexes $N(U) \to s(U)$.

**Proof.**
This is clear because the differential on $s(U)_ n = U_ n$ is $\sum (-1)^ i d^ n_ i$ and the maps $d^ n_ i$, $i < n$ are zero on $N(U_ n)$, whereas the restriction of $(-1)^ nd^ n_ n$ is the boundary map of $N(U)$ by definition.
$\square$

## 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Your email address will not be published.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.

## Comments (0)