# 范畴嵌入复形　`prop.category-into-complexes`
命题：$M \mapsto [M]$ 把范畴全忠实地嵌进复形范畴
layer 11 · 命题 · 复形与导出三角 · 同调代数
[在星图里打开](../#n=prop.category-into-complexes)　·　[节点索引](../index.md)

把整数集看成偏序范畴 $(\mathbb{Z}, \le)$。由含入 $\{0\} \hookrightarrow (\mathbb{Z},\le)$ 得到函子

$$\mathcal{C} \longrightarrow \mathcal{C}(\mathcal{C}), \qquad M \longmapsto [M], \qquad [M]^{n} := \begin{cases} M, & n = 0 \\ 0, & n \ne 0\end{cases}$$

（微分全为零）。则：
> 陈述续见 [nodes/prop.category-into-complexes.2.md](./prop.category-into-complexes.2.md)


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

refs: Le Stum, An introduction to condensed mathematics, §7.1

> 说明见 `notes/prop.category-into-complexes.md`

引用：`prop.category-into-complexes`（内容 v0.9.0）· 条款见 [LICENSE.txt](../LICENSE.txt)