Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

A number of topological invariants take the form of functors $\mathscr{T}\to\mathscr{G}$, where $\mathscr{T}$ is the category of all topological spaces and continuous functions, and $\mathscr{G}$ is the category of all groups and homomorphisms. For examples, consider the homology groups $H_{n}(X)$ or the homotopy groups $\pi_{n}(X)$. Of course, a problem with these invariants is that they are not fully faithful functors, i.e., $H_{n}(X)\cong H_{n}(Y)$ does not imply that $X$ and $Y$ are homeomorphic. The existence of a fully faithful functor $F:\mathscr{T}\to\mathscr{G}$ would imply that $\mathscr{G}$ has a subcategory $F\mathscr{T}$ equivalent to $\mathscr{T}$. This would be both rather disturbing and extremely interesting. First, it would mean that in a sense, all of topology is just a subset of group theory, which would be rather disturbing to topologists, but it would also reveal a fundamental connection between two seemingly disparate disciplines. My question is: is it possible to prove that no such functor exists? In other words, could one exhibit some categorical property that $\mathscr{T}$ posesses that $\mathscr{G}$ does not. This question can naturally be extended to other important categories, like $\mathscr{M}$, the category of all modules, or $\mathscr{R}$, the category of all rings. So in general, given arbitrary categories $\mathscr{C}$, $\mathscr{D}$, is there any natural way of showing that no fully faithful functor $F:\mathscr{C}\to\mathscr{D}$ exists, i.e., are there any nice "categorical invariants?"

EDIT: A couple people pointed out that I really ought to be discussing fully faithful functors, rather than just faithful functors. Also, I have changed the title in accordance with Martin Brandenburg's recommendation.

share|improve this question
Barring there being an actual interesting property about an embedding of say, the topological category to the category of groups IMO it's not clear there would be much interest in such a functor. Saying one impossible problem is a subset of another collection of impossible problems isn't particularly useful. Having functors that forget a lot of structure is a fundamentally useful thing, it's in a sense what we aim for since it allows actual reduction of problems to tractable problems. –  Ryan Budney Sep 30 '10 at 1:42
I think that you mean to say "fully faithful". The forgetful functor from spaces to sets is faithful, as is the "free abelian group" functor from sets to groups. –  Tom Goodwillie Sep 30 '10 at 1:53
"it would mean that in a sense, all of topology is just a subset of group theory" -- I think this is a fallacious interpretation, even if such an embedding did exist, for the reasons given by Ryan and Tom. –  Yemon Choi Sep 30 '10 at 3:36
I don't see what is so disturbing about finding a way to encode topological spaces in groups. Infinite groups can be extremely complicated. –  S. Carnahan Sep 30 '10 at 4:07
@Daniel: Very nice question. In your title you speak of equivalence, in your homologiy example you speak of isomorphic-reflecting, then of faithful functors, but I think you mean fully faithful functors. Perhaps you should also edit the title. It's about non-embeddabilty of categories. –  Martin Brandenburg Sep 30 '10 at 13:17

9 Answers 9

up vote 3 down vote accepted

One property that $Top_{cgwh}$ has (if we let $Top_{cgwh}$ be the full subcategory of $Top$ consisting of compactly generated, weakly Hausdorff spaces) that $Grp$ doesn't is cartesian closedness: the hom-set $Grp(G,H)$ is not a group. Another property that $Top$ has that $Grp$ doesn't is an initial object distinct from its final object, but these are just off the top of my head.

But thinking of these functors as being intrinsic to $Top$ is a small mistake, because $Top$ is just a presentation of the $(\infty,1)$-category of homotopy types, and these functors are really giving invariants of homotopy types.

This doesn't really answer your question, but I hope it indicates that the sort of functor you are looking for won't be anything like cohomology or homotopy.

Edit: what about the composite $Top \to Set \to Grp$ where the first is the 'underlying set' functor and the second is the 'free group' functor? This is faithful, but perhaps not very useful for topological purposes. (Edit again: I just noticed that Tom Goodwillie also pointed out a similar example)

share|improve this answer
I should just point out that given a functor Top-> Grp as you are looking for would make Top_cgwh a subcategory of Grp, but cartesian closedness and subcategories don't necessarily get along. –  David Roberts Sep 30 '10 at 6:10
Of course, saying that Top is only a presentation of the $(\infty,1)$-category of homotopy types is being a little bit unfair. Indeed, it has forgetful functors to the category of locales and the category of sets, which show that there's an even richer structure on actual topological spaces. –  Harry Gindi Sep 30 '10 at 7:39
(than just the homotopical structure, that is). –  Harry Gindi Sep 30 '10 at 7:40
Of course there is, extra structure but the functors taken as examples (pi_n, H_n) descend to the (oo,1)-category, where there are fewer morphisms, and this is one point to consider when talking about faithful functors out of Top: they can't be invariant under homotopy –  David Roberts Sep 30 '10 at 7:44
Just about the terminology: The (oo,1)-category does not have fewer morphisms, the homotopy category does. –  Peter Arndt Sep 30 '10 at 15:21

The category of pointed topological spaces has no fully faithful functor to Groups, because I can think of a pointed space that has exactly 3 automorphisms whereas there is no group with that property. (My space is non-Hausdorff and has only 10 points.)

share|improve this answer
I like that. Too often we only think of big spaces. Give finite spaces a chance. They are great fun and very versatile. –  Tim Porter Sep 30 '10 at 16:30
Nice! Here is a proof that there is no group $G$ with exactly $3$ automorphisms: Assume $Aut(G)$ has order $3$. Then $Aut(G)$ is cyclic, which implies that $G$ is abelian. Then $x \mapsto -x$ is an automorphism of order $2$, contradiction. –  Martin Brandenburg Sep 30 '10 at 16:58
Along the lines of what you suggest, there are exactly 2 groups with Aut(G) trivial (a standard homework problem), whereas there are many spaces with that property (including many finite spaces). –  Tyler Lawson Oct 1 '10 at 2:02
Or the empty set. The only other Hausdorff examples I can think of are mortifyingly big (formed by, say, starting with an interval, then gluing in endpoints to intervals at the rational points while simultaneously gluing more endpoints to the new intervals... so that eventually every rational point of an interval has an integer number of endpoints glued to it, with this integer unique for every such point). –  Tyler Lawson Oct 1 '10 at 2:55
However, one can cheat without employing such spaces and say that any embedding would have to send the empty set and the one-point set to the two groups with trivial Aut-group. Then these two groups have maps in both directions, but the empty set and the one-point set don't. –  Tyler Lawson Oct 1 '10 at 2:57

There is no full embedding of the category of all topological spaces into the category of groups because $\mathscr{T}$ is not bounded (i.e., it has no dense subcategory), while $\mathscr{G}$ is locally presentable, and assuming Vop$\rm{\check{e}}$nka's principle a category $C$ is fully embeddable into a locally presentable category iff $C$ is bounded. (All the definitions are in [AR] Adamek, Rosicky's "Locally presentale and accessible categories", and the theorem is Theorem 6.6). This argument also excludes full embeddings of $\mathscr{T}$ into any other locally presentable (or accesible) category, hence any category of models of a basic theory ([AR], Th.5.35). As a consequence, $\mathscr{T}$ is not embeddable into the category of rings, the category of modules over a ring, the category of small categories or the category of simplicial sets.

On the other hand, $\mathscr{T}$ is not bounded because (recall that a subcategory $A\subset C$ is dense if it is small and every object $c$ of $C$ is a canonical colimit of objects in $A$) for each cardinal $\lambda$ there is a space $X$ which is not discrete, but whose spaces of cardinality smaller than $\lambda$ are all discrete. So if $A$ were a dense subcategory of $\mathscr{T}$, taking $\lambda$ as the cardinality of the biggest space in $A$, the space $X$ cannot be a colimit of objects in $A$.

These spaces do not appear in most of the categories of spaces algebraic topologists are used to work with, so this argument only applies for the whole $\mathscr{T}$. For example, $\Delta$-generated spaces are bounded and even locally presentable.

Finally, as an example, let me point out that it is possible to fully embedd the category of $T_1$ $\Delta$-generated spaces into the category of small categories (which are monoids with several objects), in particular, you can do the same with the category of $CW$-complexes.

share|improve this answer

I'm going to answer (for my own limited version of the word "answer") the actual question and ignore the motivation (essentially for the reasons given in the comments). And to explain my limited version of the word "answer", what I'm going to do is explain how I attacked this problem in a recent paper where I wanted to show that two categories were not equivalent. In the absence of a proper answer, this might give someone some ideas on how to do it (or else in the comments to this "answer" I'll get told of a much better way of doing what I did and then MO will get another paper to its citation list as I have to rewrite my paper to take that into account). The paper in question is: Comparative Smootheology and this particular bit is section 9.

As background, the categories that I wanted to compare were the different versions of the categories of "smooth stuff". I'd already shown that these weren't equivalent as extensions of $\operatorname{Diff}$, but I wanted to show that there was absolutely no equivalence whatsoever. I did it basically by showing that there were certain "categorical invariants" that had to be preserved under an equivalence.

  1. The first was the forgetful functor to $\operatorname{Set}$. In each case, this functor was represented by a terminal object and so would be preserved (up to natural isomorphism) by any equivalence. Using this, I could show that if my categories were equivalent, then they were equivalent as constructs (ie concrete categories over $\operatorname{Set}$.

  2. The above meant that I could look at the fibre categories over a particular set. This allowed me to separate my categories (there were five that I was considering) into two by looking at the cardinality of one particular fibre category (in this case, over $\{0,1\}$: three categories had fibre of size $2$, two of size $4$. Did I claim that this was a sophisticated method?).

  3. Next, I looked at endomorphism monoids of particular objects. These are preserved under equivalences. I found that in one of my categories, there was a single object (up to isomorphism within the category, which could be fixed by considering the underlying set) with a particular endomorphism monoid (details: in the category of Frölicher spaces, only $\mathbb{R}$ has endomorphism monoid $C^\infty(\mathbb{R},\mathbb{R})$). Extending my vision slightly, by considering the endomorphism monoids, I could find three objects $X \to Y \to Z$ with the same underlying set (in this case $\mathbb{R}$) such that the endomorphism monoid of each contained $C^\infty(\mathbb{R},\mathbb{R})$. By "find", I mean that I could categorically point to them.

  4. By considering similar objects in the other categories, I could show that any equivalence of categories had to preserve this particular object. In other words, in all of the categories I can categorically point to $\mathbb{R}$.

  5. At this point, my work is almost done. Once I can categorically point to $\mathbb{R}$, I can categorically point to any smooth manifold. That means that an equivalence between any two of my categories must preserve the subcategory of smooth manifolds (more carefully said: any equivalence must itself be equivalent to one which preserves ...). But earlier in the paper, I'd shown that no such existed.

  6. Ta-da!

If anyone is feeling a little underwhelmed by that, I'll quote from the paper:

In this section we have proceeded on a mixture of general results and case-by-case analysis. It is entirely possible that a fuller analysis of the general structure would lead to an elimination of the more ad-hoc aspects, but as the goal was to analyse the specific examples of these categories, where a simple ad-hoc argument sufficed we deemed it more appropriate to give it than to search for what could be a more complicated but more general result.

In other words, if anyone is feeling a little underwhelmed, I agree with you!

share|improve this answer
I like your answer. It's difficult to prove that two abstract categories are not isomorphic, you have to find many specific invariants. On the other hand, your techniques can only be used when we want to show that Top and Grp are not equivalent. Daniel asks if Top is equivalent to a full subcategory of Grp. –  Martin Brandenburg Sep 30 '10 at 13:08

There is no full functor $Top \to Grp$ at all, since between every two groups there is a morphism, which is not true for spaces.

Perhaps a more interesting question is if there exists a fully faithful functor $hCW_* \to Grp^\omega$, where $hCW_*$ is the homotopy category of pointed CW-complexes and $Grp^\omega$ is the category of sequences of groups. Note that Whitehead's lemma implies that $\pi_{\bullet} : hCW_* \to Grp^\omega$ reflects isomorphisms.

share|improve this answer
If we exclude the empty set from being a topological space, then given any two spaces $X$ and $Y$, can we not simply map all of $X$ to any point in $Y$? –  Charles Staats Sep 30 '10 at 15:54
Yes, but who cares about the category of all nonempty topological spaces? –  Martin Brandenburg Sep 30 '10 at 16:09
+1 to Martin's comment, but we could think about pointed spaces. See my somewhat silly answer. –  Tom Goodwillie Sep 30 '10 at 16:24

I'm sure some people can give more refined answers than this one, but you can get basic invariants of a category by comparing isomorphism types with connected components of the nerve. Informally, you are examining what changes (on a coarse level) when you invert all of the arrows. For example, topological spaces always have a map to a point, so any functor from $\textbf{Top}$ to a groupoid will take all objects to the same isomorphism type. In particular, groupoids cannot distinguish any pair of topological spaces (or any pair of nonisomorphic objects in another category, if the objects have a zigzag of arrows between them).

share|improve this answer

There are some well-known results that could be thought of as positive answers in this direction, at least up to homotopy.

First of all, the Kan-Thurston theorem says that every connected CW complex can be recovered up to homotopy type by performing the plus-construction on a certain discrete group built from the space. The construction was cleaned up a bit in "The topology of discrete groups" by Baumslag-Dyer-Heller (J. Pure Appl. Algebra 16 (1980), no. 1, 1--47). From the mathscinet review: "[The construction] is interpreted as giving an equivalence between the homotopy category of connected pointed CW complexes and a category of fractions of the category of group eipmorphisms with perfect kernels."

Also, there's a theorem of McDuff which says that every path connected space X has the weak homotopy type of the classifying space of a discrete monoid (essentially a monoid of configurations of points in X).

share|improve this answer

There is no full* functor from pointed spaces, or its homotopy category, to groups, because there is no full* functor from pointed sets to groups. Proof: Let $F$ be such a functor. Denote by $n$ a based set with $n$ elements. Then the group $F(1)$ must be trivial because it has only one endomorphism; the group $F(2)$ must be of order two because it has exactly two endomorphisms; the group $F(3)$ must have have exactly two automorphisms, making it abelian, and on the other hand it must have exactly three morphisms from $F(2)$. Contradiction.

We could ask about the homotopy category of path-connected based spaces (which I suppose is just as off-putting, Martin, as the category of non-empty spaces, but let's go on anyway). This has a full functor from groups, so if it had a full functor to groups then composing we would get a full functor from groups to groups which is not a equivalence of categories. Sounds impossible, but I haven't got an argument.

EDIT: * fully faithful

share|improve this answer
You also used faithfulness of $F$ ("exactly three morphisms"). –  Martin Brandenburg Oct 1 '10 at 8:37
I may be wrong but I think that what used to be called "fully faithful" or "full and faithful" is nowadays called "full". Anyway, that's what I meant. –  Tom Goodwillie Oct 1 '10 at 10:33
I don't think I've ever heard "full" used to include faithfulness as well (I work mainly in "pure" category theory and categorical logic). Is it used that way in some other fields? –  Peter LeFanu Lumsdaine Oct 2 '10 at 18:48

I think No (for reasons of category theory):

Let $1\in Set$ the final object (the one-point set) considered as discrete topological space , then $\mathcal{T}(1, X)\cong |X|$ (where $|X|$ is the support set of $X$), then $Gr(F(1), F(1)\cong \mathcal{T}(1, 1)$ has only one element, then $1_{F(1)}$ is the trivial morphism (constant on the identity) then $F(1)$ is the trivial (only-identity) group. Then for any group $G$ the set $Gr(F(1), G)$ has only one element, then for any topological space $X$ follow that $|X|\cong \mathcal{T}(1, X)\cong Gr(F(1), F(X)$ has only one element (absurd).

Please, forgive me for my bad english.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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