Lemma 21.9.4. Let $\mathcal{C}$ be a category. Let $\mathcal{U} = \{ f_ i : U_ i \to U\} _{i \in I}$ be a family of morphisms with fixed target such that all fibre products $U_{i_0} \times _ U \ldots \times _ U U_{i_ p}$ exist in $\mathcal{C}$. The chain complex $\mathbf{Z}_{\mathcal{U}, \bullet }$ of presheaves of Lemma 21.9.3 above is exact in positive degrees, i.e., the homology presheaves $H_ i(\mathbf{Z}_{\mathcal{U}, \bullet })$ are zero for $i > 0$.
Proof. Let $V$ be an object of $\mathcal{C}$. We have to show that the chain complex of abelian groups $\mathbf{Z}_{\mathcal{U}, \bullet }(V)$ is exact in degrees $> 0$. This is the complex
For any morphism $\varphi : V \to U$ denote $\mathop{\mathrm{Mor}}\nolimits _\varphi (V, U_ i) = \{ \varphi _ i : V \to U_ i \mid f_ i \circ \varphi _ i = \varphi \} $. We will use a similar notation for $\mathop{\mathrm{Mor}}\nolimits _\varphi (V, U_{i_0} \times _ U \ldots \times _ U U_{i_ p})$. Note that composing with the various projection maps between the fibred products $U_{i_0} \times _ U \ldots \times _ U U_{i_ p}$ preserves these morphism sets. Hence we see that the complex above is the same as the complex
Next, we make the remark that we have
Using this and the fact that $\mathbf{Z}[A] \oplus \mathbf{Z}[B] = \mathbf{Z}[A \amalg B]$ we see that the complex becomes
Finally, on setting $S_\varphi = \coprod _{i \in I} \mathop{\mathrm{Mor}}\nolimits _\varphi (V, U_ i)$ we see that we get
Thus we have simplified our task. Namely, it suffices to show that for any nonempty set $S$ the (extended) complex of free abelian groups
is exact in all degrees. To see this fix an element $s \in S$, and use the homotopy
with obvious notations. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)