# 伴随的复合　`prop.adjunction-composition`
命题：$F_{1} \dashv G_{1}$、$F_{2} \dashv G_{2}$ $\implies$ $F_{2}F_{1} \dashv G_{1}G_{2}$
layer 10 · 命题 · 伴随与反射 · 范畴论
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若 $F_{1} \dashv G_{1}$（$\mathcal{C} \to \mathcal{D}$）与 $F_{2} \dashv G_{2}$（$\mathcal{D} \to \mathcal{E}$），则

$$F_{2} \circ F_{1} \;\dashv\; G_{1} \circ G_{2}.$$

## 为什么成立（入边，证明在 proofs/）
- …另有入边，续页见 [nodes/prop.adjunction-composition.2.md](./prop.adjunction-composition.2.md)

refs: Le Stum, An introduction to condensed mathematics, Exercise 1.59

> 说明见 `notes/prop.adjunction-composition.md`

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