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This question is related to this one: Which limits are preserved by a reflective left-adjoint?

(And in fact may be seen as a special case, but I think it merits its own question).

Suppose that $C$ is a reflective subcategory of $D$ and that coproducts in $D$ commute with finite limits. Notice that coproducts computed in $C$ are not necessarily the same as computing them in $D$; one may have to apply the reflector afterwards. Does it however follow that coproducts in $C$ commute with finite limits? Edit: No. See Todd's answer below. If not, under what assumptions will this be nonetheless true?

Edit: Obviously this is true if the reflector is left-exact. I am interested in conditions weaker than this.

Subquestion: Is there anything that can be said in the special situation where $C$ is the category of free-algebras for a monad on $D$ and the adjunction is the obvious one?

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A simple example where finite limits do not commute with coproducts is where $D$ is the category of functors $T \to Set$ for just about any Lawvere theory $T$ that comes to mind, say the theory of groups, and $C$ is the category of product-preserving functors $T \to Set$. Clearly $C$ is a full subcategory of $D$ and the inclusion preserves limits and has a left adjoint, but in $C$ (which is equivalent to the category of groups), finite limits do not commute with coproducts.

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  • $\begingroup$ Ah, good point Todd. I forgot that algebras for Lawvere theories are reflective. So, no, apparently this is not true in general. I still would like to know when it IS true however. (It's funny you mention Lawvere theories since my example involves them). $\endgroup$
    – David Carchedi
    Commented May 9, 2012 at 2:00
  • $\begingroup$ By the way, a notable exception to your family of counterexamples is the Lawvere theory of commutative $mathbb{R}$-algebras. Can we see what is special about this example at the level of Lawvere theories, that makes coproducts commute with finite limits? (By the way, my reflective subcategory is a subcategory of commutative $\mathbb{R}$-algebras as it turns out). $\endgroup$
    – David Carchedi
    Commented May 9, 2012 at 2:18
  • $\begingroup$ @David: The tensor product with a fixed commutative algebra preserves finite products of modules since they are also finite coproducts, so it preserves finite products of commutative algebras which can be calculated as products in modules. Now we just need to check that it preserves equalizers which can be calculated as the kernel of the difference of the maps, since our fixed commutative algebra is \emph{flat} (since we are working over the reals) tensoring with our algebra preserves kernels. $\endgroup$
    – Justin Noel
    Commented May 9, 2012 at 19:04

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