# 切片上的上同调就是原上同调　`prop.cohomology-slice-identification`
命题：$H^{n}\bigl(\mathbf{1}_{X}, j_{X}^{-1}M^{\bullet}\bigr) = H^{n}(X, M^{\bullet})$
layer 19 · 命题 · 拓扑斯 · 范畴论
[在星图里打开](../#n=prop.cohomology-slice-identification)　·　[节点索引](../index.md)

设 $\mathcal{T}$ 是拓扑斯、$X \in \mathcal{T}$、$M^{\bullet}$ 是阿贝尔群复形。则
> 陈述续见 [nodes/prop.cohomology-slice-identification.2.md](./prop.cohomology-slice-identification.2.md)


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

refs: Le Stum, An introduction to condensed mathematics, Exercise 7.59

> 说明见 `notes/prop.cohomology-slice-identification.md`

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