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What forgetful functors are equivalences?

The motivation here is understanding when some part of a structure can be 'safely' forgotten, even if remembering it might make our lives easier.

There is some ambiguity as to what constitutes 'a forgetful functor', but I am open to any answers that are 'obvious' forgetful functors or where the interpretation is justified.

I don't have any nontrivial examples (the forgetful functor from the $EM$-category of the trivial monad on a category), and I am open to arguments that no nontrivial examples exist. Any relevant input is appreciated.

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    $\begingroup$ @SamHopkins That functor is fully faithful, but it's not an equivalence. $\endgroup$ Commented Apr 30, 2022 at 0:24
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    $\begingroup$ Right, of course. But it is an example of a forgetful functor that "doesn't forget anything." $\endgroup$ Commented Apr 30, 2022 at 0:25
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    $\begingroup$ @SamHopkins I don't think it does constitute an example of not forgetting anything in the sense that the OP's asking about: while in a sense it doesn't forget anything about individual objects (being fully faithful), it forgets a lot about the global nature of the source category. $\endgroup$ Commented Apr 30, 2022 at 0:26
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    $\begingroup$ @AlecRhea, so you're meaning to ask about something like "redundant structure"? :) $\endgroup$ Commented Apr 30, 2022 at 0:54
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    $\begingroup$ I suppose one type of example comes about whenever an object specified by a wishlist of nice properties turns out to be uniquely determined by those properties. For example you could consider the category of Riemannian manifolds equipped with a torsion-free connection compatible with the metric. Then the forgetful functor to Riemannian manifolds is an isomorphism of categories, since the Levi-Civita connection is the only such connection (but of course this fact is nontrivial). Probably more interesting examples would be those that given an equivalence of categories rather than an isomorphism. $\endgroup$
    – user164898
    Commented Apr 30, 2022 at 1:31

15 Answers 15

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The functor that takes a simplicial abelian group to its associated chain complex is arguably a "forgetful" functor. By the Dold-Kan theorem it induces an equivalence of categories.

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    $\begingroup$ That's a great answer. $\endgroup$ Commented Apr 30, 2022 at 6:23
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    $\begingroup$ This is exactly the kind of answer I was hoping for, thank you Gregory. $\endgroup$
    – Alec Rhea
    Commented Apr 30, 2022 at 10:26
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    $\begingroup$ To me, this functor doesn't look 'forgetful' ― there's really an interesting construction going on, not just the omission of part of the data or part of the axioms. $\endgroup$ Commented Apr 30, 2022 at 22:55
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    $\begingroup$ I agree it is a bit of a stretch. The output of this functor appears to have less structure than the input (though it turns out not really). So it "feels" like a functor that forgets information. But perhaps it is too elaborate to be really called a forgetful functor. Maybe it boils down to this. Consider taking the kernel as a functor from the arrow category of abelian groups to abelian groups. One could argue that this is a forgetful functor. If you accept this, then the Dold-Kan functor is "just" forgetting a bunch of maps, taking the kernel of a bunch of maps, and only remembering one map $\endgroup$ Commented May 1, 2022 at 10:00
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By the positive solution to Hilbert's fifth problem (as well as the theorem of Cartan that continuous homomorphisms between Lie groups are automatically smooth), the forgetful functor from the category of Lie groups (with smooth homomorphisms as the morphisms) to the category of locally Euclidean topological groups (with continuous homomorphisms as the morphisms) is an equivalence.

I would imagine that many rigidity theorems (e.g., Mostow rigidity) can also be reformulated as an assertion that something resembling a forgetful functor is in fact an equivalence, but I have not worked this out carefully. (For instance, Gromov's quasi-isometric rigidity program can be sort of interpreted as the question of measuring the extent to which the forgetful functor from finitely generated groups (with virtual homomorphisms rather than global homomorphisms as morphisms) to metric spaces (with quasi-isometries as the isomorphisms) fails to be an equivalence. The Ribe program to analyze the local geometry of Banach spaces via their metric structure (up to uniformly bicontinuous maps) is in a similar spirit, and inspired by the Ribe rigidity theorem, though a little less easy to describe explicitly in terms of forgetful functors.)

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    $\begingroup$ This is very interesting, thank you — I initially asked this question as a lark, but it’s looking like the ‘nontrivial’ examples of this phenomenon are actually interesting theorems. $\endgroup$
    – Alec Rhea
    Commented Apr 30, 2022 at 16:58
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The forgetful functor from group objects in abelian groups to abelian groups (just forget the extra group structure) is an equivalence.

More generally there are lots of "idempotent" constructions, such as "taking abelian group objects in" or "taking pointed objects in" that give examples like this. These are the "uninteresting" ones in that they are very general.

An example not of this type is given by duality data. Let's say you have symmetric monoidal category $C$. Inside there you have dualizable objects, and they form a full subgroupoid $C^{dbl}$. But a dualizable object can also be seen as a bunch of duality data: a tuple $(x,y, x\otimes y \to 1)$ for example. This type of data forms a groupoid $DDat(C)$ and the forgetful functor $DDat(C) \to C^{dbl}$ is an equivalence.

Another example that happens a lot in ordinary category theory is when forgetting some higher dimensional coherence data ends up not being necessary. For example, forgetting from "very structured monoids" to simply monoids is an equivalence, where a "very structured monoid" is a functor $\Delta^{op}\to C$ satisfying the Segal conditions - similarly with "very structured commutative monoids", or more generally algebras or commutative algebras.

Similarly one can forget from very structured monoidal categories to monoidal categories - although that is an equivalence only if one views both as $(2,1)$-categories. But I think the above shows that this phenomenon occurs even in ordinary categorical situations.

Let me give a final example: there are lots of criteria like "to preserve colimits, it suffices to preserve coequalizers and coproducts". This can be seen as a forgetful equivalence: from the category whose objects are cocomplete categories and arrows are colimit preserving functors to the category with the same objects but functors are functors which preserve only certain types of colimits. These statements are easy, but it is still an example of this form.

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  • $\begingroup$ Small typo: I don't think you mean for $C^{dbl}$ to be full, just the subcategory of dualizable objects and isomorphisms thereof. $\endgroup$
    – Brian Shin
    Commented May 1, 2022 at 2:49
  • $\begingroup$ @BrianShin : I think that's what's typically called "full subgroupoid", but I have to say I'm not 100% sure $\endgroup$ Commented May 1, 2022 at 7:56
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My favorite such functor is the forgetful functor from "$C^\infty$ manifolds with real coefficients" (defined as spaces with a sheaf of $\mathbb{R}$-algebras locally isomorphic to the sheaf $C^\infty_\mathbb{R}$ on the space $\mathbb{R}^n$) to $C^\infty$ manifolds with complex coefficients (define as spaces with a sheaf of $\mathbb{C}$-algebras locally isomorphic to the sheaf $C^\infty_{\mathbb{C}}$ on $\mathbb{R}^n$). This equivalence fails to hold in the supergeometric world (as well as in various other extensions of the theory of real manifolds).

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The forgetful functor from abelian varieties to pointed schemes is fully faithful, (hence an equivalence onto its essential image). That is, a pointed map of schemes between abelian varieties automatically preserves the multiplication.

(Maybe some assumptions about the base need to be made for this to be true? I'm not sure. At any rate, I'm pretty sure I've been told that this is at least true over $\mathbb C$.)

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    $\begingroup$ Do you have a reference for that? $\endgroup$
    – Z. M
    Commented Apr 30, 2022 at 18:53
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    $\begingroup$ @Z.M This is discussed eg on the nlab page. $\endgroup$
    – Tim Campion
    Commented May 1, 2022 at 3:51
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    $\begingroup$ This is true for abelian schemes over any base scheme $S$; see for instance Bhatt's notes, Corollary 2.3. (Archived version on wayback machine because the original will probably disappear at some point in the not-so-far future.) $\endgroup$ Commented May 1, 2022 at 14:44
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    $\begingroup$ Other references for the statement: Mumford's Abelian Varieties, section 4, corollary 1 (for abelian varieties over a field); Mumford, Fogarty, and Kirwan's Geometric Invariant Theory, corollary 6.4 (for abelian schemes over any base scheme). $\endgroup$ Commented May 3, 2022 at 5:19
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    $\begingroup$ And another reference: Hindry and Silverman's Diophantine Geometry, corollary A.7.1.2 (for abelian varieties over a field). $\endgroup$ Commented May 3, 2022 at 5:26
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An example from a pedagogical perspective: the category of based vector spaces, i.e. vector spaces with a chosen basis, and morphisms matrices, is equivalent to the category of vector spaces by the functor which forgets the basis. Similarly: the category of homotopy types with a choice of CW complexes realizing the type, and morphisms homotopy classes of continuous maps. Generally, this happens any time a choice of extra structure is always possible and doesn’t change the morphisms, which of course is just rephrasing the request that the forgetful functor be essentially surjective and fully faithful!

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One reason to believe that this phenomenon is ubiquitous yet not so interesting is the following construction:

Given an equivalence of categories $F \colon \mathscr C \stackrel\sim\to \mathscr D$, define the category of triples $(c,d,f)$ of an object $c \in \mathscr C$, an object $d \in \mathscr D$, and an isomorphism $f \colon F(c) \stackrel\sim\to d$. It has forgetful projections to $\mathscr C$ and $\mathscr D$ that are both equivalences.

For an example of this phenomenon in action, consider the equivalence between Boolean rings and Boolean lattices. It's not really forgetful in the sense that you need to define some new structure going both ways. However, we can define a category of Boolean ring lattices $(A,+,\cdot,0,1,\wedge,\vee)$ expressing both sets of axioms and the relations between them (ok, in this case $\cdot$ and $\wedge$ are literally the same operation). Then both forgetful functors are equivalences.

Likewise, there is an equivalence between algebraic lattices $(L,\wedge,\vee)$ and posets $(P,\leq)$ with finite products and coproducts, and one can consider the object $(L,\wedge,\vee,\leq)$ remembering both. Then the forgetful functors are equivalences.

(In both examples, we only work with $f \colon F(c) \stackrel\sim\to d$ that are the identity on some fixed object $F(c) = d$.)

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  • $\begingroup$ So if I understand you correctly, you suspect that many instances of ‘forgetful equivalences’ will be generated from another ‘non-forgetful equivalence’ via this construction? $\endgroup$
    – Alec Rhea
    Commented Apr 30, 2022 at 10:22
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    $\begingroup$ I guess so. For me, forgetful means that you take the sets of data and axioms, and remember only a subset of each (this is a procedure in logic more than category theory). For instance, the Dold–Kan example as given now is not strictly speaking forgetful, because you have to make a new construction (the normalised chain complex of a simplicial object). You can make it forgetful by adding this redundant data to the language of simplicial objects, but then you're doing exactly what my answer describes. $\endgroup$ Commented Apr 30, 2022 at 10:53
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    $\begingroup$ Ah, but I am more convinced by some of the other answers, such as Terry Tao's or Tim Campion's. $\endgroup$ Commented Apr 30, 2022 at 17:07
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Group objects in the category of smooth manifolds (aka Lie groups) are the same thing as monoid objects in the category of smooth manifolds whose underlying monoid is a group.

Said in other words: the inverse map, when it exists, is automatically (continuous and) smooth.

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An important example from type theory/categorical logic: the forgetful functor from categories with families (or variants) to categories with attributes (or variants).

Briefly, both are category-based structures encoding the structure of a dependent type theory (and hence the structure needed to interpret such a theory). A category with families consists of a category of contexts, a presheaf of types over contexts, and a further presheaf of terms indexed over types, together with an operation of context extension satisfying a certain universal property.

The universal property forces terms to correspond to sections of certain projection maps. Categories with attributes take advantage of that, including just the category of contexts, the presheaf of types, the extension operation, and the projection maps.

So the forgetful functor is indeed an equivalence, and genuinely not an isomorphism: Given a CwF, if you forget to the underlying CwA, you can recover original the presheaf of terms just up to isomorphism.

Each notion has some advantages. CwA’s are a little “tighter” and easier to set up, since they involve less data; but taking the preheaf of terms as primary, in CwF’s, simplifies some later constructions, and allows very fruitful analyses and generalisations.

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There is an obstruction to nontrivial examples. With strong enough metatheory (with enough global choice), equivalent categories of structures (in which the isomorphism class of an object is a proper class) are isomorphic. I think the argument is not very standard, so I will summarize it here.

Let $F$ be an equivalence between $C$ and $D$. For each isomorphism class of objects in $C$, fix a bijection from that isomorphism class to the isomorphism class containing its direct image under $F$. Define $G$ from the objects of $C$ to the objects of $D$ as follows:

For each object $A$, let $G(A)$ be the image of $A$ under the fixed bijection $g_{[A]}$ from the isomorphism class of $[A]$ to the isomorphism class $[F(A)]$.

For each object $A$, choose an isomorphism $h_A$ from $F(A)$ to $G(A)$.

For each arrow $s$ from $A$ to $B$, let $G(s)$ be the arrow $h_B.F(s).(h_A)^{-1}$ from $G(A)$ to $G(B)$.

Since $F$ is an equivalence, one can check that $G$ is an isomorphism of categories.

Therefore, if a forgetful functor (between categories of structures) is an equivalence, then the categories must be isomorphic.

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    $\begingroup$ Note however that “the categories are isomorphic” does NOT imply “the forgetful functor is an isomorphism”. Indeed, in the most interesting examples, it won’t be an isomorphism — essentially, when the “extra data” that is forgotten is only unique up to isomorphism, not unique on the nose. $\endgroup$ Commented May 1, 2022 at 11:28
  • $\begingroup$ Yes, of course, that's why I have written "the categories must be isomorphic", not "the forgetful functor must be an isomorphism". $\endgroup$ Commented May 1, 2022 at 11:43
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    $\begingroup$ Certainly, I wasn’t taking issue with anything you wrote (I think your answer is a nice observation) — just adding a caveat, since I’ve often seen people misinterpret such statements. $\endgroup$ Commented May 1, 2022 at 20:38
  • $\begingroup$ Sure. The argument is quite general: if an equivalence of categories F is given such that the isomorphism class of A is equipotent with the isomorphism class of F(A), then the categories are isomorphic (under global choice). $\endgroup$ Commented May 2, 2022 at 0:34
  • $\begingroup$ But in particular, following Peter's comment, I think this is not an obstruction to "nontrivial" examples. As you point out, this has little to do with categories of structures and just about size issues. Unlike in other areas where you have both equivalences and isomorphisms, I think having an isomorphism of categories is such an unnatural notion that an equivalence being liftable to an isomorphism does not make the equivalence trivial $\endgroup$ Commented May 2, 2022 at 8:12
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Let $\mathcal{F}$ be the forgetful functor from the category of compact uniform spaces (the compact uniform spaces are the complete and totally bounded uniform spaces) to the category of compact proximity spaces. Let $\mathcal{G}$ be the forgetful functor from the category of compact proximity spaces to the category of compact Hausdorff spaces. Then the functors $\mathcal{F},\mathcal{ G},\mathcal{G}\circ\mathcal{F}$ are isomorphisms, so $\mathcal{F},\mathcal{G},\mathcal{G}\circ\mathcal{F}$ are equivalences of categories as well.

More generally, if $(X,\mathcal{U}),(Y,\mathcal{V})$ are uniform spaces, $(X,\mathcal{U})$ is compact, and $f:X\rightarrow Y$ is continuous, then $f$ is automatically uniformly continuous.

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What about the classical theorem that the natural transformation $\eta$ witnessing the adjointess of two functions $F, G$:

$$ \alpha_{X, Y} : Hom(FX, Y) \simeq Hom(X, GY) $$

is determined by the values $$\epsilon_X:= \alpha_{X, FX} (1_{FX}) : X \to GFX$$ $$\eta_Y:= \alpha_{GY, Y}^{-1}(1_{GY}) : FGY \to Y$$

I always found this argument clever and simple, despite it is very basic.

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The forgetful functor from Z-modules to abelian groups (which forgets the scalar multiplication) is an isomorphism.

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$\mathrm{C}^*$-algebras $\to$ Banach algebras that happen to admit an involution satisfying the $\mathrm{C}^*$-algebra axioms.

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    $\begingroup$ Can you describe the category on the right in more detail, in particular what the morphisms are? I am having trouble fleshing out this example without making it tautologous. $\endgroup$
    – Yemon Choi
    Commented Apr 30, 2022 at 14:05
  • $\begingroup$ For that matter: is the category on the left what used to be called (the category of) $B^*$-algebras, i.e. Banach star-algebras with an involution satisfying the Cstar condition? If Cstar algebras here mean closed star subalgebras of some B(H), then what are the morphism? $\endgroup$
    – Yemon Choi
    Commented Apr 30, 2022 at 14:06
  • $\begingroup$ @YemonChoi: you're right, I've been a bit sloppy. I just remembered that if a Banach algebra has an involution that makes it into a C*-algebra, then such involution is unique. I was assuming the morphisms on both sides are just continuous homomorphisms. Perhaps it's a bit tautologous (many things in category theory are tautologous in sense though..). If you know some relevant functional analysis to make my answer more interesting you're welcome to do so. :) $\endgroup$
    – Qfwfq
    Commented Apr 30, 2022 at 16:27
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    $\begingroup$ One of my points is that you haven't specified if your homomorphisms between the Banach algebras that happen to have suitable involutions are supposed to be star homomorphisms. In the category of Cstar algebras you want the morphisms to be star homomorphisms. So I think this example might need to be taken back to the drawing board. $\endgroup$
    – Yemon Choi
    Commented Apr 30, 2022 at 18:16
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    $\begingroup$ If your work in the respective categories where morphisms are required to have norm at most 1 then we might get something like your original claim, via the Vidav-Palmer theorem; but I am currently occupied with other tasks and haven't sat down to work out the details. $\endgroup$
    – Yemon Choi
    Commented Apr 30, 2022 at 18:21
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For $k$ big enough, any $C^k$ map between finite-dimensional vector spaces which commutes with scalar multiplication is a homomorphism.

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