# 层化不改变导出的整体截面　`prop.cohomology-presheaf-H`
命题：$\Gamma\bigl(X, R\mathcal{H}(M^{\bullet})\bigr) = R\Gamma(X, M^{\bullet})$，从而 $\Gamma\bigl(X, \mathcal{H}^{n}(M^{\bullet})\bigr) = H^{n}(X, M^{\bullet})$
layer 18 · 命题 · 层上同调 · 同调代数
[在星图里打开](../#n=prop.cohomology-presheaf-H)　·　[节点索引](../index.md)

设 $X \in \mathcal{C}$、$M^{\bullet}$ 是阿贝尔层复形，$\mathcal{H}$ 记「把阿贝尔层当预层看」的含入函子。则
> 陈述续见 [nodes/prop.cohomology-presheaf-H.2.md](./prop.cohomology-presheaf-H.2.md)


## 为什么成立（入边，证明在 proofs/）
- …另有入边，续页见 [nodes/prop.cohomology-presheaf-H.3.md](./prop.cohomology-presheaf-H.3.md)

refs: Le Stum, An introduction to condensed mathematics, Exercise 7.42

> 说明见 `notes/prop.cohomology-presheaf-H.md`

引用：`prop.cohomology-presheaf-H`（内容 v0.9.0）· 条款见 [LICENSE.txt](../LICENSE.txt)