Examples of categories over $\bf{Set}$ such that the forgetful functor is conservative include the "algebraic" categories of groups, rings, modules, monoids, etc., but does not include the "higher-order" categories of topological spaces, smooth manifolds, schemes, etc. or some "relational" categories like posets.

Is there any intrinsic characterization of the difference in behavior here? Do we know that certain classes of categories always have conservative forgetful functors to $\bf{Set}$?

I'd be particularly interested in a model-theoretic description (i.e. something like the fact that the first examples are categories of models of an algebraic first order theory while the later ones are not), but I'd also be interested in a description in terms of categorical properties.

EDIT: As pointed out in the answers, it's often possible to give a conservative faithful functor to $\bf{Set}$. However, these aren't the "underlying set" forgetful functors; they include extra information. Maybe the question becomes more interesting if we require the functor to be "canonical". This doesn't make much sense in terms of categories alone, but perhaps it does in terms of model theory: given a (perhaps higher-order) theory $\mathbb{T}$, when is the forgetful functor $\text{MOD-}\mathbb{T}(\bf{Set}) \rightarrow \bf{Set}$ conservative?

  • 1
    $\begingroup$ Won't something like $\mathbf{Top}\to \mathbf{Set}$, $(X,\tau)\mapsto \mathcal P(\tau)$, $f\colon(X,\tau)\to(X',\tau')\;\mapsto\;S\mapsto\{U\in\tau': f^{-1}(U)\in S\}$ work for $\mathbf{Top}$? $\endgroup$ – Hagen von Eitzen Apr 6 '15 at 11:34
  • 1
    $\begingroup$ Is this functor faithful? It seems like something silly could happen for the indiscrete topology. (Although maybe we could just have the product of this functor and the usual forgetful functor) $\endgroup$ – Dorebell Apr 8 '15 at 5:01
  • $\begingroup$ The edit asks a very good question. $\endgroup$ – goblin Jan 5 '16 at 17:20
  • $\begingroup$ nice.............+1 $\endgroup$ – Bhaskara-III Jan 11 '16 at 15:57

A conservative functor $\mathsf{Top} \to \mathsf{Set}$ has been described in the comments. In a similar vein, for schemes we can take $X \mapsto |X| \amalg \mathcal{P}(\mathrm{Stalks}(X))$. This should also work for manifolds. Actually, for manifolds the much more formal $X \mapsto \coprod_{Y ~ \mathrm{conn}} \mathrm{Hom}(Y,X)$ works too, where the coproduct is over connected $Y$. So there are more categories conservative over $\mathsf{Set}$ than you might think.

It might be interesting to ask which categories admit a representable conservative functor to $\mathsf{Set}$. Or at least "familialy representable": a set of objects $S_\alpha$ such that $\coprod_\alpha \mathrm{Hom}(S_\alpha,-)$ is conservative is called a strong generator for a category. Categories with strong generators are too general to admit a good classification, but I'm pretty sure that topological spaces and schemes don't have strong generators (although manifolds do: the connected manifolds). Also, see the nlab article for other sorts of "generators" (the nlab prefers the term "separator"). Generating hypotheses are important auxiliary hypotheses in theorems which often rule out categories of a "topological" nature.

One question which has received attention in the categorical literature is when a category admits a faithful functor to $\mathsf{Set}$, or is "concrete". Famously, the homotopy category of topological spaces does not. Necessary and sufficient conditions are known characterizing when a category is concrete -- if the category has finite limits, then the condition is equivalent to being regularly well-powered (i.e. every object has a small set of regular subobjects). But this question doesn't really get at the distinction you want: the usual functor $\mathsf{Top} \to \mathsf{Set}$ is faithful, for example.

Stronger notions which may be related to the distinction you want include accessible categories, topological categories, and algebraic categories.

As to the edit, if $\mathbb{T}$ is a first-order theory, I can think of several reasonable categories of models to consider:

  • $\mathrm{Mod}(\mathbb{T})$, the category of models and homomorphisms. In this case, it seems to me that one easy criterion for the forgetful functor to be conservative is that there be no relation symbols (except equality), but only function symbols (including constants) in the language. The converse -- if the forgetful functor is conservative then $\mathbb{T}$ can be axiomatized over a language with no relation symbols -- might be true, I don't know. EDIT Actually, this converse probably fails. For example, let $\mathbb{T}$ be the theory with two binary relations $R,S$ and the axiom $\forall x,y \, R(x,y) \Leftrightarrow \neg S(x,y)$. Then the forgetful functor is conservative, but it seems very unlikely that this theory can be axiomatized in a signature with no relations.

  • $\mathrm{Elem}(\mathbb{T})$, the category of models and elementary embeddings. In this case, it seems to me that the forgetful functor to $\mathsf{Set}$ is always conservative.

  • For just about any reasonable class of formulas $\Phi$, the category $\Phi-\mathrm{Elem}(\mathbb{T})$, the category of models and homomorphisms which preserve the satisfaction of formulas in $\Phi$. $\mathrm{Mod}(\mathbb{T})$ is the case when $\Phi$ is just the atomic formulas (those built up from variables, function symbols, and relation symbols, but no logical connectives or quantifiers). As long as $\Phi$ contains the atomic formulas and their negations (so that the morphisms of $\mathrm{Mod}(\mathbb{T})$ are all embeddings or at least strong homomorphisms, I believe that the forgetful functor to $\mathsf{Set}$ is conservative.

As for higher-order logic, I really don't know because I'm not familiar enough with it.

| cite | improve this answer | |
  • $\begingroup$ Thanks for the answer - I never thought to be more creative with the choice of functor. Is the question in the edit asking something more interesting? $\endgroup$ – Dorebell Apr 8 '15 at 5:03
  • 1
    $\begingroup$ Well, if you restrict to categories of models of first-order theories, I think the underlying set functor is always conservative. Trying to force the forgetful functor to be "canonical" is part of the idea behind asking it to be representable or a coproduct of representables. More broadly, just asking that it be a small colimit of representables is already reasonably restrictive. $\endgroup$ – tcamps Apr 8 '15 at 12:46
  • $\begingroup$ Is this true if the category allows maps that are arbitrary model homomorphisms as opposed to elementary embeddings? It seems like you may run into trouble with certain relations (although it is of course true if the theory is algebraic) $\endgroup$ – Dorebell Apr 9 '15 at 13:24
  • $\begingroup$ Thanks for catching my error-- the category of simple graphs is a counterexample (taking the underlying set to be the vertices); but I suspect it doesn't matter if you use elementary embeddings or homomorphisms. But if you're talking about the models of a first-order theory, you always have an accessible category (whether you take homomorphisms or elementary embeddings as morphisms), so there will be a dense generator, which is a fortiori a strong generator, so there exists a familially representable conservative functor to $\mathsf{Set}$. $\endgroup$ – tcamps Apr 9 '15 at 13:58
  • $\begingroup$ Possibly a dumb question, but is being the category of models of a monosorted first-order theory equivalent to being an accessible category? The nLab article is a little vague on this point. $\endgroup$ – goblin Jan 6 '16 at 15:53

The nicest kind of forgetful functors are those which are monadic, which is a reasonable categorical description of categories of models of some sort of theory. Monadic functors are always conservative. And a nice characterization is known of categories monadic over $\text{Set}$ (this is Borceux, Handbook of Categorical Algebra, Volume 2, Theorem 4.4.5):

Theorem: A category $C$ is monadic over $\text{Set}$ iff it

  • has finite limits,
  • is (Barr) exact, and
  • has a regular generator $P$, meaning an object $P$ such that the coproduct $\coprod_X P$ exists for every set $X$ and such that, for every object $Y \in C$, the natural map $\coprod_{f : P \to Y} P \to Y$ is a regular epimorphism.

With these hypotheses, the monadic forgetful functor is $\text{Hom}(P, -)$.

For example, $\text{Top}$ is not Barr exact, so it is not monadic over $\text{Set}$ with respect to any forgetful functor. On the other hand, the category of compact Hausdorff spaces is monadic over $\text{Set}$; the corresponding monad is the ultrafilter monad, which you can think of as an infinitary algebraic theory whose operations are given by taking limits wrt ultrafilters.

| cite | improve this answer | |
  • 2
    $\begingroup$ You have forgotten one hypothesis : the regular generator $P$ must be regular projective, i.e. the functor $Hom(P,\_)$ must preserve regular epimorphisms. $\endgroup$ – Arnaud D. May 22 '17 at 14:28

Freyd proved in "On the concreteness of certain categories" that if $C$ is locally small than there is a conservative functor $C \to \text{Set}$.

| cite | improve this answer | |
  • $\begingroup$ Um, are you sure you mean "locally small" as opposed to "small" here? That paper's hard to find. $\endgroup$ – Kevin Arlin Feb 4 '17 at 15:40
  • $\begingroup$ Yes, I should be sure. $\endgroup$ – Ivan Di Liberti Feb 4 '17 at 16:11
  • $\begingroup$ I was actually interested in whether you can offer any more evidence for what strikes me as an extraordinary claim, since as I said I can't see the paper. $\endgroup$ – Kevin Arlin Feb 4 '17 at 16:13
  • $\begingroup$ I will write you an email right now. $\endgroup$ – Ivan Di Liberti Feb 4 '17 at 16:14
  • 4
    $\begingroup$ Cool! It turns out to be elementary. Define $T: C \to \mathsf{Set}$ by $T(A) = \amalg_X C(X,A) / \sim$ where $f \sim f'$ if for all $g: A \to B$, $gf$ is iso iff $gf'$ is iso. Note that any two maps which are not split mono are identified in $T(A)$ (and so $T(A)$ is small if $C$ is locally small). Define $T_\ast: C \to \mathsf{Set}$ by setting $T_\ast(A) = T(A)$ unless every map into $A$ is split mono, in which case a disjoint basepoint is added. It's straightforward to show that $T_\ast$ reflects the property of being split mono and also the property of being split epi, so is conservative. $\endgroup$ – tcamps Feb 5 '17 at 9:03

The following seems true:

Proposition. Let $\sigma$ denote a first-order signature with only function symbols (i.e. no relations). Let $\mathsf{T}$ denote a first-order theory in the language of $\sigma$. Then $\mathrm{Mod}(\mathsf{T})$ is conservatively concrete over $\mathbf{Set}$.

I've made this community wiki; can anyone provide a proof?

| cite | improve this answer | |
  • 1
    $\begingroup$ The proof is pretty trivial. It's just the statement that if a bijection $f$ turns an $n$-ary operation $P$ into an $n$-ary operation $P'$, then $f^{-1}$ turns $P'$ into $P$. $\endgroup$ – Eric Wofsey Jan 11 '16 at 21:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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