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?

  • $\begingroup$ How trivial of a solution would you accept? For example, the category with a single object (any set of your choice) and one arrow. $\endgroup$ – Zev Chonoles Apr 25 '13 at 0:45
  • 3
    $\begingroup$ 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. $\endgroup$ – Zhen Lin Apr 25 '13 at 7:12
  • 2
    $\begingroup$ 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. $\endgroup$ – Martin Brandenburg Apr 25 '13 at 8:03
  • $\begingroup$ 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 $S$... $\endgroup$ – Nex Nov 25 '15 at 13:10
  • $\begingroup$ 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? $\endgroup$ – Giorgio Mossa Jul 7 '16 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.