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

- All Categories 2.3K
- Chat 499
- Study Groups 17
- Petri Nets 8
- Epidemiology 3
- Leaf Modeling 1
- Review Sections 9
- MIT 2020: Programming with Categories 51
- MIT 2020: Lectures 20
- MIT 2020: Exercises 25
- MIT 2019: Applied Category Theory 339
- MIT 2019: Lectures 79
- MIT 2019: Exercises 149
- MIT 2019: Chat 50
- UCR ACT Seminar 4
- General 67
- Azimuth Code Project 110
- Statistical methods 3
- Drafts 2
- Math Syntax Demos 15
- Wiki - Latest Changes 3
- Strategy 113
- Azimuth Project 1.1K
- - Spam 1
- News and Information 147
- Azimuth Blog 149
- - Conventions and Policies 21
- - Questions 43
- Azimuth Wiki 707

Options

## Comments

By definition of product (Def. 3.81), there are morphisms\(\pi_1:x\times y\to x\), \(\pi_2:x\times y\to 2\). So, \(x\leq(x\times y)\) and \(y\leq(x\times y)\). This means that \(x\times y\) is an upper bound for \(x\) and \(y\). By the universal property of products, any \(Z\) such that there are morphisms \(f:Z\to x\) and \(g:Z\to y\) has a unique morphism into \(x\times y\). Thus, \(x\times y\) is a least upper bound for \(x\) and \(y\), i.e., it equals \(x\wedge y\).

`By definition of product (Def. 3.81), there are morphisms\\(\pi_1:x\times y\to x\\), \\(\pi_2:x\times y\to 2\\). So, \\(x\leq(x\times y)\\) and \\(y\leq(x\times y)\\). This means that \\(x\times y\\) is an upper bound for \\(x\\) and \\(y\\). By the universal property of products, any \\(Z\\) such that there are morphisms \\(f:Z\to x\\) and \\(g:Z\to y\\) has a unique morphism into \\(x\times y\\). Thus, \\(x\times y\\) is a least upper bound for \\(x\\) and \\(y\\), i.e., it equals \\(x\wedge y\\).`