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 18
- Petri Nets 9
- 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 708

Options

*Uniqueness of universal objects.*

Recall the definitions of terminal object and product from Chapter 3 in the notes.

(a) Show that if \(t\) and \(t'\) are both terminal objects in a category, then \(t\) and \(t'\) are isomorphic.

(b) Let \(a\) and \(b\) be objects of a category. Show that if \(p\) and \(p'\) are both products of \(a\) and \(b\), then they are isomorphic.

(c) Discuss the similarities between your two proofs. Could the same idea be used to show that any two initial objects are isomorphic?

## Comments

(a) Let us define the functions \( f : t \rightarrow t' \) and \( g : t' \rightarrow t \) Their composition gives us \( g \circ f : t \rightarrow t \) and \( f \circ g : t' \rightarrow t' \) Since there is only one morphism from any \( a \) to \( t \) or \( t' \) then these two compositions must necessarily be the identity function.

(b) Since \( p, p' \) are both products of \( a, b \) then there is a unique morphism from \( c \) to them ( \( h, h' \) ). If t we have two morphisms \( \alpha : p \rightarrow p' \) and \( \beta : p' \rightarrow p \) . Since \( h, h' \) are unique, we have:

$$ \alpha \circ h : c \rightarrow p' = h' $$ $$ \beta \circ h' : c \rightarrow p = h $$ $$ \alpha \circ \beta \circ h' = h' $$ $$ \beta \circ \alpha \circ h = h $$ And our only choice is for \( \alpha, \beta \) be the identity functions.

(c) both proofs used the uniqueness of morphisms and the commuting diagrams. I think we could follow the same idea for the initial objects, since there is only one morphism from \( 0 \rightarrow \ \).

`(a) Let us define the functions \\( f : t \rightarrow t' \\) and \\( g : t' \rightarrow t \\) Their composition gives us \\( g \circ f : t \rightarrow t \\) and \\( f \circ g : t' \rightarrow t' \\) Since there is only one morphism from any \\( a \\) to \\( t \\) or \\( t' \\) then these two compositions must necessarily be the identity function. (b) Since \\( p, p' \\) are both products of \\( a, b \\) then there is a unique morphism from \\( c \\) to them ( \\( h, h' \\) ). If t we have two morphisms \\( \alpha : p \rightarrow p' \\) and \\( \beta : p' \rightarrow p \\) . Since \\( h, h' \\) are unique, we have: $$ \alpha \circ h : c \rightarrow p' = h' $$ $$ \beta \circ h' : c \rightarrow p = h $$ $$ \alpha \circ \beta \circ h' = h' $$ $$ \beta \circ \alpha \circ h = h $$ And our only choice is for \\( \alpha, \beta \\) be the identity functions. (c) both proofs used the uniqueness of morphisms and the commuting diagrams. I think we could follow the same idea for the initial objects, since there is only one morphism from \\( 0 \rightarrow \ \\).`