The Stacks project

Lemma 14.27.1. Let $\mathcal{A}$ be an additive category. Let $a, b : U \to V$ be morphisms of simplicial objects of $\mathcal{A}$. If $a$, $b$ are homotopic, then $s(a), s(b) : s(U) \to s(V)$ are homotopic maps of chain complexes. If $\mathcal{A}$ is abelian, then also $N(a), N(b) : N(U) \to N(V)$ are homotopic maps of chain complexes.

Proof. We may choose a sequence $a = a_0, a_1, \ldots , a_ n = b$ of morphisms from $U$ to $V$ such that for each $i = 1, \ldots , n$ either there is a homotopy from $a_ i$ to $a_{i - 1}$ or there is a homotopy from $a_{i - 1}$ to $a_ i$. The calculation above shows that in this case either $s(a_ i)$ is homotopic to $s(a_{i - 1})$ as a map of chain complexes or $s(a_{i - 1})$ is homotopic to $s(a_ i)$ as a map of chain complexes. Of course, these things are equivalent and moreover being homotopic is an equivalence relation on the set of maps of chain complexes, see Homology, Section 12.13. This proves that $s(a)$ and $s(b)$ are homotopic as maps of chain complexes.

Next, we turn to $N(a)$ and $N(b)$. It follows from Lemma 14.23.6 that $N(a)$, $N(b)$ are compositions

\[ N(U) \to s(U) \to s(V) \to N(V) \]

where we use $s(a)$, $s(b)$ in the middle. Hence the assertion follows from Homology, Lemma 12.13.1. $\square$


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Your email address will not be published.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 019S. Beware of the difference between the letter 'O' and the digit '0'.