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.
Best Answer
As k.stm says in the comments, usually a more general thing is true: these categories $C$ are equipped with forgetful / underlying set functors $U : C \to \text{Set}$ which tend to have a left adjoint, the "free" functor $F : \text{Set} \to C$. Whenever this is true, it follows that $U$ preserves all limits, not just products.
Sometimes, but more rarely, $U$ will also have a right adjoint, which will give a "cofree" functor. Whenever this is true, it follows that $U$ preserves all colimits, not just coproducts. This happens, for example, when $C = \text{Top}$: here the left adjoint equips a set with the discrete topology and the right adjoint equips a set with the indiscrete topology. But it doesn't happen for, say, groups or rings.
So one way to reformulate your question is:
A rough answer is that we can expect "free" structures whenever the structure is described by operations satisfying equational axioms, since then we can build free objects by applying all possible operations modulo all axioms. On the other hand, operations also make it difficult for the forgetful functor to preserve coproducts, since in a coproduct of two structures you can apply operations to elements of both structures, so you'll usually get something bigger than the disjoint union.
(Dually, you should expect "cofree" structures whenever the structure is described by "co-operations," and this does in fact happen: for example, the forgetful functor from coalgebras to vector spaces has a right adjoint but not a left adjoint, called the cofree coalgebra.)
A more precise answer would invoke, say, the machinery of Lawvere theories, which among other things has the benefit of also telling you exactly what colimits you can expect these forgetful functors $U$ to preserve. This is a long story so I don't want to get into it unless you feel like it really answers your question, but the gist is that Lawvere theories present familiar structures like groups, rings, and modules in a way that fundamentally uses finite products, but nothing else. You can deduce from this that the forgetful functor $U$ preserves (and in fact creates) any limits or colimits that commute with finite products in $\text{Set}$. Every limit commutes with finite products, and the colimits that commute with finite products in $\text{Set}$ are precisely the sifted colimits. These include, for example, increasing unions, which is an abstract way to see why the set-theoretic increasing union of a sequence of groups is still a group, and the same with groups replaced by rings, modules, etc.