It looks like you're new here. If you want to get involved, click one of these buttons!

- All Categories 2.2K
- Applied Category Theory Course 354
- Applied Category Theory Seminar 4
- Exercises 149
- Discussion Groups 49
- How to Use MathJax 15
- Chat 480
- Azimuth Code Project 108
- News and Information 145
- Azimuth Blog 149
- Azimuth Forum 29
- Azimuth Project 189
- - Strategy 108
- - Conventions and Policies 21
- - Questions 43
- Azimuth Wiki 711
- - Latest Changes 701
- - - Action 14
- - - Biodiversity 8
- - - Books 2
- - - Carbon 9
- - - Computational methods 38
- - - Climate 53
- - - Earth science 23
- - - Ecology 43
- - - Energy 29
- - - Experiments 30
- - - Geoengineering 0
- - - Mathematical methods 69
- - - Meta 9
- - - Methodology 16
- - - Natural resources 7
- - - Oceans 4
- - - Organizations 34
- - - People 6
- - - Publishing 4
- - - Reports 3
- - - Software 21
- - - Statistical methods 2
- - - Sustainability 4
- - - Things to do 2
- - - Visualisation 1
- General 39

Options

Complete the proof

- Show, that if f is left adjoint to g then for any q ∈ Q, we have f (g(q)) ≤ q.
- Show that If 1.6 holds, then for any p ∈ P and q ∈ Q, if p ≤ 1(q) then f (p) ≤ q.

**Galois connection**
$$
f(p) \le q \iff p \le g(q) \\
\text{Where g is right-adjoint to f and f is left-adjoint to g}
$$
**1.6**
$$
\text{For every } p \in P \:\&\: q \in Q \\
p \le g(f(p)) \:\&\: f(g(q)) \le q
$$

## Comments

Since the first inequality holds by reflexivity, the desired relation must also hold.

`1. By adjointness of f,g: g(q)\\(\leq\\)g(q)\\(\\Leftrightarrow\\)f(g(q))\\(\leq\\)q. Since the first inequality holds by reflexivity, the desired relation must also hold. 2. Assume p≤g(q). By monotonicity of f, f(p)≤f(g(q)). By 1.6 and transitivity, f(p)≤q, as was to be proved.`