# 高次上同调预层没有 Čech 上同调　`prop.cech-higher-cohomology-presheaf`
命题：$n \ne 0 \implies \check{H}^{p}\bigl(H^{n}(M)\bigr) = 0$
layer 19 · 命题 · 层上同调 · 同调代数
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设 $M$ 是 site 上的阿贝尔层、$n \ne 0$，并把 $H^{n}(M)$ 记作预层 $U \mapsto H^{n}(U, M)$。则
> 陈述续见 [nodes/prop.cech-higher-cohomology-presheaf.2.md](./prop.cech-higher-cohomology-presheaf.2.md)


## 它能推出什么 / 谁在用它
- ⇒ [`prop.cech-h1-equals-h1`](../nodes/prop.cech-h1-equals-h1.md) 一次 Čech 就是一次上同调

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

refs: Le Stum, An introduction to condensed mathematics, Exercise 7.45

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

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