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Question 2.4 - Uniqueness of universal objects

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?


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  • 1.
    edited January 23

    (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 \( 0 \rightarrow \ \).

    Comment Source:(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 \ \\).
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