# 加法函子与复形构造交换　`prop.additive-complex-construction`
命题：$F(C(\mathcal{A})) \cong C(F(\mathcal{A}))$
layer 11 · 命题 · 复形与导出三角 · 同调代数
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加法函子 $F : \mathcal{A} \to \mathcal{A}'$ 逐项作用给出函子

$$C(F) : C(\mathcal{A}) \longrightarrow C(\mathcal{A}'), \qquad (K^{n}, d^{n}) \longmapsto \bigl(FK^{n},\ Fd^{n}\bigr),$$
> 陈述续见 [nodes/prop.additive-complex-construction.2.md](./prop.additive-complex-construction.2.md)


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

refs: Le Stum, An introduction to condensed mathematics, Exercise 7.27

> 说明见 `notes/prop.additive-complex-construction.md`

引用：`prop.additive-complex-construction`（内容 v0.9.0）· 条款见 [LICENSE.txt](../LICENSE.txt)