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# Question 2.4 - Uniqueness of universal objects

edited January 2020

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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edited January 2020

(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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