TITLE: Why is the cartesian product so categorically robust?
QUESTION [41 upvotes]: The major "broad/natural" categories I encounter in daily life are: sets, groups, topological spaces, smooth manifolds, vector spaces over a fixed field $k$, $k$-schemes, rings, $A$-algebras for a fixed commutative ring $A$, and $A$-modules. I think that about covers it. (I am using the word "broad/natural" to mean things that might have occurred to me as categories when I first learned what a category was, i.e. to distinguish them from the more "clever" and "restricted" categories one uses to articulate specific theorems, such as the category of finite sets equipped with a continuous action of a fixed profinite group, or the category of open subsets of a topological space.) 
Today I found myself really struck by the fact that in almost all of these categories, the categorical product exists and its underlying set is the cartesian product, whereas almost all the categories have different underlying sets for coproducts. In other words, in almost all these categories, taking products commutes with the forgetful functor, whereas in almost none of them does taking coproducts commute with the forgetful functor $F$.
Specifically:
Groups: $\prod$ is cartesian product; $\coprod$ is free product. (Only $\prod$ commutes with $F$.) 
Topological spaces and smooth manifolds: $\prod$ is cartesian product; $\coprod$ is disjoint union. (Both commute with $F$.)
$k$-vector spaces, $A$-modules: $\prod$ is cartesian product; $\coprod$ is direct sum. (Only $\prod$ commutes with $F$.)
$A$-algebras: $\prod$ is the cartesian product; $\coprod$ is the tensor product over $A$. (Only $\prod$ commutes with $F$.) Rings are the special case $A=\mathbb{Z}$.
The only case, among those I listed above, where the product $\prod$ does not commute with the $F$ is the category of $k$-schemes.
My question is:

Can you help me think about why this is happening? Why is it so much more likely, at least for the "broad/natural" categories one encounters in real life, for the forgetful functor to commute with taking products than with taking coproducts?

Thanks in advance.

REPLY [3 votes]: A simple answer that covers most of the purely algebraic examples is that  any class of structures that can be axiomatised by Horn clauses gives rise to a concrete category whose forgetful functor commutes with products. (In another question Concrete category with non-standard products, I use the term concrete category with standard products for this property.)
The axioms for topological spaces and manifolds of various kinds can be cooked up to look like Horn clauses in some kind of infinitary logic, so the standard model theory proof about structures defined by Horn clauses may well adapt to work for them as well. How this ties up with Lawvere theories as mentioned by Qiaochu Yuan is an interesting question.