# Why is the cartesian product so categorically robust?

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?

• Well, those are concrete categories, so the fact that you already have the maps available from the category of sets goes a long way towards it. – Tobias Kildetoft May 29 '15 at 20:05
• I don’t know all of these categories, but – at least as far as I know – for most of them the corresponding forgetful functor has a left adjoint, thus is continuous. Morally speaking, I’d guess that so many of these forgetful functors have left adjoints because mathematicians (for historic reasons alone) tend to build stuff from sets and that already suggests a forgetful functor as well as an left adjoint for it. – k.stm May 29 '15 at 20:17
• This reminds me of this answer: math.stackexchange.com/a/25460/2614 – Bruno Stonek May 29 '15 at 22:25

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:

Why do the forgetful functors $U : C \to \text{Set}$ we write down tend to have left adjoints, but not right adjoints? Equivalently, why are there usually free structures, but not usually cofree structures?

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.

• Instead of Lawvere theories, I would like to throw in monads. Monadic functors preserve all limits, and create those colimits which the underlying functor of the monad preserves. – Martin Brandenburg Jun 3 '15 at 8:07
• @Martin: that's certainly true, but then one is led to ask the question "why are many of the familiar concrete categories we care about monadic over $\text{Set}$?" One answer is to invoke the monadicity theorem, but this already requires that we know that the forgetful functor has a left adjoint, and in particular that it preserves limits. Another is to describe a machine that constructs monadic adjunctions out of other data like Lawvere theories or operads, which is why I took that route instead. – Qiaochu Yuan Jun 3 '15 at 16:58
• There is no real difference, since you also would have to check if your favorite forgetful functor is "Lawveric". – Martin Brandenburg Jun 3 '15 at 18:23
• @Martin: no, I think there is a difference. Categories like groups, rings, and modules already come to us as categories of models of a Lawvere theory: I'm not applying a recognition principle when I say this (which would be something like "monadic + the monad preserves sifted colimits"), I'm directly looking at the way these categories are defined. – Qiaochu Yuan Jun 4 '15 at 5:11

The question can be rephrased to: why does the forgetful functor so often preserve limits? (Because in all of the cases you mention equalisers are preserved too.) This property of the forgetful functor is forced by abstract nonsense (representable functors are continuous) and by the fact that forgetful functors very often are representable. As for why the latter is the case, I don't think it's that unexpected: representability of a forgetful functor means that there exists an object $R$ such that the elements of any object $C$ correspond (nicely) to morphisms $R → C$. In geometry one expects there is indeed such a space (a point, although note how this fails for schemes, where products are not preserved), and in algebra this is exactly the property of the free algebraic structure on one element (eg. $\mathbb Z$ for $\mathrm{Grp}$ or $k[X]$ for $k$-algebras).

In fact, often more is true: the forgetful functor $U : \mathcal C → \mathrm{Set}$ has a left adjoint F, and is in particular representable (for the one element set $1$ and an arbitrary object $C$ adjointness boils down to $\mathrm{Hom}(F1, C) ≅ \mathrm{Hom}(1, UC) ≅ UC$, ie. $U$ is represented by the free object on one element, as I mentioned for algebraic structures). Adjointness however is not necessary so I think representability is the more important and more intuitive reason for continuity.

• If $C$ has coproducts (which is true in all of the OP's examples) then the forgetful functor being representable is equivalent to it having a left adjoint. – Qiaochu Yuan May 29 '15 at 22:21
• @QiaochuYuan: Thanks, I should've pointed that out. You don't always have them though (eg. complete lattices or I guess manifolds depending on how you define them) and I do think representability is the more basic and accessible reason. – user54748 May 29 '15 at 22:35
• That's fair. And it is worth noticing that there isn't an obvious analogue of representability which ensures that a functor $C \to \text{Set}$ preserves colimits. – Qiaochu Yuan May 29 '15 at 22:38
• @QiaochuYuan: Indeed, but I don't know if preservation of colimits can be treated by such general arguments anyway (my impression is that cofree functors for example are rather exceptional, although I'm not familiar with enough "categories of spaces" to give a better estimation). – user54748 May 29 '15 at 22:53

Here are some counterexamples.

The category $\mathsf{Ban}_1$ (objects: Banach spaces over $\mathbb{C}$; morphisms: short linear maps) has products, but the forgetful functor $U : \mathsf{Ban}_1 \to \mathsf{Set}$ only prefers finite products. The product of an (infinite) family of Banach spaces $(V_i,\lVert - \rVert_i)$ is given by $(P,\lVert - \rVert)$, where $$P = \bigl\{v \in \prod_{i \in I} V_i : \lVert v \rVert := \sup_{i \in I} \lVert v_i \rVert_i < \infty\bigr\}.$$ But there is a better forgetful functor, the unit ball functor $B_1 : \mathsf{Ban}_1 \to \mathsf{Set}$. This is actually isomorphic to $\hom_{\mathsf{Ban}_1}(\mathbb{C},-)$ and thus preserves all limits for general reasons.

For the category of schemes $\mathsf{Sch}$, the forgetful functor $U : \mathsf{Sch} \to \mathsf{Set}$ does not preserve the terminal object, i.e. the empty product. One can say exactly in how it does not preserve fiber products: For morphisms $f : X \to S$, $g : Y \to S$, the natural map $U(X \times_S Y) \to U(X) \times_{U(S)} U(Y)$ is surjective, and the fiber of $(x,y) \in U(X) \times_{U(S)} U(Y)$ lying over $s \in U(S)$ identifies with $\mathrm{Spec}(\kappa(x) \otimes_{\kappa(s)} \kappa(y))$.

• What do you think the forgetful functor $\mathbf{Rel} \to \mathbf{Set}$ is? There is one, but it's not identity-on-objects. – Zhen Lin Jun 3 '15 at 8:38
• You are right, I have deleted that paragraph. The faithful "power set" functor $\mathsf{Rel} \to \mathsf{Set}$ preserves products, right? – Martin Brandenburg Jun 3 '15 at 8:45
• Right. It's representable, after all. – Zhen Lin Jun 3 '15 at 8:47
• Ah, this is just $\hom_{\mathsf{Rel}}(\{\star\},-)$. Thank you. – Martin Brandenburg Jun 3 '15 at 8:48

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.

• Lawvere theories are the category-theoretic version of algebraic theories, which are a special case of universal Horn theories. This suggests that we ask: what is the categorical version of universal Horn theories? Unfortunately, the relevant nLab is downright lame, but I'll let you know if anything pops up. – goblin Jun 24 '15 at 16:59