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In Rel, both products and coproducts amount to taking a disjoint union of sets. Is there a category whose object class can be interpreted as the class of all sets, in which both products and coproducts amount to taking Cartesian products, in the classical sense?

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How trivial of a solution would you accept? For example, the category with a single object (any set of your choice) and one arrow. – Zev Chonoles Apr 25 '13 at 0:45
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Easy: let $\mathcal{C}$ be the category whose objects "are" sets, and whose morphisms $X \to Y$ are the group homomorphisms from the free abelian group generated by $X$ to the free abelian group generated by $Y$. This has the required property. – Zhen Lin Apr 25 '13 at 7:12
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So you claim that $\hom(F(X \times Y),F(Z)) \cong \hom(F(X),F(Z)) \times \hom(F(Y),F(Z))$ and $\hom(F(X),F(Y \times Z)) \cong \hom(F(X),F(Z)) \times \hom(F(Y),F(Z))$? Neither is clear to me. – Martin Brandenburg Apr 25 '13 at 8:03
    
Yes take $\mathbb{C}$ to be the category with objects sets and with a unique isomorphism between any pair of objects. This type of category has many strange properties amongst which one easily sees that the product of $X$ and $Y$ is any set $S$ (in particular $X\times Y$ will do), and similarly the coproduct of $X$ and $Y$ is again any set $S$... – Nex Nov 25 '15 at 13:10
    
I have trouble understanding what exactly are you looking for: for how you asked the question I would say that $\mathbf{Set}$, the category of set and functions, satisfies your requirement...... so what are you exactly looking for? – Giorgio Mossa Jul 7 at 10:30

In the category of Abelian groups:

If $A$ and $B$ are abelian groups, the product $A\times B$ together with $\iota_A:a \longmapsto (a,0)$ and $\iota_B:b \longmapsto (0,b)$ is a co-product. Since if $f:A \longrightarrow C$ and $B \longrightarrow C$ are morphisms of abelian groups, then $h:A\times B \longrightarrow C$, defined by $(a,b)\longmapsto f(a)+g(b)$, is the unique homomorphism making the required (coproduct) diagram commute.

This is only true for finite-products, but hopefully it answers the question.

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