Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $\mathcal{D}$ be a full subcategory of a category $\mathcal{C}$. Let $X, Y$ be objects of $\mathcal{D}$. Suppose a product $X\times Y$ exists in $\mathcal{C}$. Suppose $X\times Y$ is not isomorphic to any object of $\mathcal{D}$. Can we conclude that a product of $X$ and $Y$ does not exist in $\mathcal{D}$? If not, any counter-example?

share|improve this question
    
Taking C to be the category of finite abelian groups and D the subcat. of cyclic groups of prime order you can find an example, I think. –  Mariano Suárez-Alvarez Dec 1 '12 at 23:44
    
Another quick example is for instance ${\cal D}=\textbf{Fields}$ and ${\cal C}=\textbf{CommRings}$. The product of two rings can never be a field, but some products exist in the category of fields. –  Marc Olschok Dec 4 '12 at 18:13

2 Answers 2

up vote 6 down vote accepted

Two counter-examples: one quite well-known and important, and one more elementary.

The important example: compactly generated (Hausdorff) topological spaces (briefly, k-spaces). The full subcategory of Top of these, kHaus, is a very nice category of spaces — better-behaved in many ways than Top itself.

Anyhow, the ordinary topological product of two k-spaces may not itself be a k-space; but it has an associated space (sometimes called its “k-ification”) which, while not a product in Top, is a product in kHaus.

The more elementary example is with coproducts not products, but sticking “op” on the categories involved makes it an example with products. Here, consider Abelian groups, AbGp, as a subcategory of all groups, Grp. The coproduct of two groups $A, B$ in Gp is their free product $A \ast B$. For Abelian $A$, $B$, the free product will usually not be Abelian: for instance, $\mathbb{Z} \ast \mathbb{Z}$ is the free group on two generators. However, they do also have a coproduct in AbGp, given by $A \oplus B$.


There’s a nice explanation for both these examples, and indeed for all the natural examples of this situation that I can think of.

Suppose D is a reflective subcategory of C, i.e. the inclusion functor has a left adjoint. (The Abelianisation functor Gp $\to$ AbGp is a typical example.) Then coproducts in D — more generally, any colimits — can be computed by taking the colimit in C and applying the reflector to the result. So, for instance, $A \oplus B$ is the Abelianisation of $A \ast B$. Checking that this recipe works whenever the colimits exist in C is a short diagram-chase — it’s essentially the fact that left adjoints preserve colimits.

Dually, if D is co-reflective — the inclusion has a right adjoint — then products (and arbitrary limits) can be computed in a similar fashion.

Reflective and co-reflective categories arise very often in nature; so they give lots of good examples where the (co)product in the ambient category doesn’t lie in the subcategory, but the subcategory has its own product nevertheless.

share|improve this answer

No. There's no reason for the inclusion of $D$ into $C$ to preserve products. Peter LeFanu Lumsdain has given some nice examples.

However, one situation in which the inclusion of $D$ into $C$ is guaranteed to preserve products is when it has a left adjoint (in which case the inclusion in fact preserves all limits). In this case, $D$ is said to be a reflective subcategory, and then the desired conclusion holds. For example:

  • Compact Hausdorff spaces are a reflective subcategory of topological spaces (the adjoint is Stone-Čech compactification).
  • Abelian groups are a reflective subcategory of groups (the adjoint is abelianization).
  • Sheaves are a reflective subcategory of presheaves (the adjoint is sheafification).
share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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