# 推前的复合　`prop.derived-pushforward-composition`
命题：$Rf'_{*} \circ Rf_{*} = R(f' \circ f)_{*}$
layer 18 · 命题 · 层上同调 · 同调代数
[在星图里打开](../#n=prop.derived-pushforward-composition)　·　[节点索引](../index.md)

设 $\mathcal{T} \xrightarrow{\ f\ } \mathcal{T}' \xrightarrow{\ f'\ } \mathcal{T}''$ 是拓扑斯态射的序列。则

$$Rf'_{*} \circ Rf_{*} \;=\; R(f' \circ f)_{*},$$
> 陈述续见 [nodes/prop.derived-pushforward-composition.2.md](./prop.derived-pushforward-composition.2.md)


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

refs: Le Stum, An introduction to condensed mathematics, Exercise 7.55

> 说明见 `notes/prop.derived-pushforward-composition.md`

引用：`prop.derived-pushforward-composition`（内容 v0.9.0）· 条款见 [LICENSE.txt](../LICENSE.txt)