# Does there exist a complete algebraic invariant of the homotopy type of a finite CW-complex?

Let $$\mathrm{Cell}$$ be the homotopy category of finite cell complexes. The main motive of my question

Is it true that for any algebraic category $$A$$ there is no fully faithful functor $$F: \mathrm{Cell} \to A$$?

The answer may depend on what exactly is meant by algebraic categories. The following version seems to me the most natural and interesting, but comments on any other versions are welcome.

By algebraic category here I mean the category of all models of an algebraic theory, where an algebraic theory is given by the following data:

• a finite set of carriers (i.e. sorts of elements)
• a finite set of operations returning a tuple over a tuple, the scope of which is determined by a set of equalities between operations (including the empty number of equalities, of course)
• a finite set of identities imposed on operations

Example: category of small categories

• Carriers: $$\mathrm{Ob}$$, $$\mathrm{Mor}$$
• Operations: $$\mathrm{dom}, \mathrm{cod}, \mathrm{id}, \circ$$ (the latter is defined on those pairs of morphisms for which dom = cod)
• Identities: $$(f \circ g) \circ h = f \circ (g \circ h)$$ $$\mathrm{dom}(f \circ g) = \mathrm{dom}~f,\;\; \mathrm{cod}(f \circ g) = \mathrm{cod}~g$$ $$\mathrm{dom} (\mathrm{id}_A) = A, \;\; \mathrm{cod} (\mathrm{id}_A) = A$$ $$\mathrm{id}_{\mathrm{dom} f} \circ f = f,\;\; f \circ \mathrm{id}_{\mathrm{cod} f} = f$$

In this example, all operations returned one element, the ability to return tuples was not used (also, of course, natural operations can occur in identities: projections and direct products of morphisms). This notion differs from the finitary algebraic theory and Lover's theories. It doesn't even seem to fall into what is called generalized algebraic theory because of the possibility of operations not being everywhere defined. However, I think I've seen something like this somewhere on nlab, but I can't find it again.

P.S. Of course, it is better not to use the composition symbol (in order to use it to write identities) and write all operations in a single syntax, but in this example I could not resist and used the traditional notation.

• diliberti.github.io/Talk/Golem.pdf Apr 10, 2022 at 22:59
• The notion of “algebraic” you’re looking for is probably essentially algebraic theories. For set-based models, they are also equivalent to generalised algebraic theories. E.g., to view categories as a GAT, change “a set of morphisms” to “for any objects $x,y \in C_0$, a set $C_1(x,y)$”, and then take composition to be defined on 5-tuples $(x,y,z,f,g)$ where $x,y,z \in C_0$, $f \in C_1(x,y)$, $g \in C_1(y,z)$ (no equalities needed in this specification!) Apr 11, 2022 at 6:46
• @IvanDiLiberti: Surely that should be an answer, not a comment? :-) Apr 11, 2022 at 6:47
• @AchimKrause : it's not really correct : there is no faithfull functor from the homotopy category of spaces to the category of Kan complexes, algebraic or not. Apr 11, 2022 at 14:07
• Ah yes, sorry. I was giving an answer "in spirit" to the title of the question, but of course it doesn't apply to the precise question asked. Apr 11, 2022 at 16:26

Following the Peter's suggestion, I'll turn my comment in an answer.

In general by algebraic category one means variety in the sense of universal algebra, or category of models of a (possibly multisorted) Lawvere theory. Yet, if we allow for infinitary operations, one could say that an algebraic category is the category of algebras for a Set-monad. Personally, I do not stand on any normative side, and I prefer to have an intuitive notion of what an algebraic category is.

Yet, for all the notions I can come up with, any algebraic category $$\mathcal{A}$$ has a faithful and conservative functor into $$\text{Set}$$, $$\mathcal{A} \to \text{Set}.$$

Because $$\text{Set}$$ is itself a (trivial) algebraic category, one gets to the following conclusion.

Conclusion. A category $$\mathcal{H}$$ has a faithful/conservative functor into an algebraic category if and only if it has a faithful/conservative functor into $$\text{Set}$$.

Having a faithful functor into the category of sets is called concreteness in category theory. The main paper on the topic are due to Peter Freyd.

• Freyd, On the concreteness of certain categories. 1969.
• Freyd, Homotopy is not concrete. 1970.
• Freyd, Concreteness. 1973.

Freyd's main results (cfr. the papers) are the following.

Thm 1 (Freyd). Any locally small category has a conservative functor into Set.

Thm 2 (Freyd). A locally small category has a faithful functor into Set if and only if it verifies the Isbell condition.

It's a bit hard to phrase the Isbell condition, thus I will just state the lazy Isbell criterion which gives a sufficient condition that sets the correct flavour.

Thm 3 (Lazy Freyd). Let C be a category with finite limits. C is concrete if and only if it is regular well-powered.

In his paper, Freyd shows that the Homotopy category of spaces cannot be concrete.

In 2016 I was doing my master, and I was fascinated precisely by the very motivation of this question, i.e. the possibility of using algebraic gadgets to distinguish topological structures. While I found Freyd's answer absolutely beautiful, I was wondering whether variations of Ho(Top) could still be concrete.

In 2017, this line of thoughts led to a collaboration with Fosco Loregian.

DL and Loregian. Homotopical Algebra is not concrete. 2018.

In the paper we should that a vast majority of homotopy categories cannot be concrete, due to the existence of Eilenberg-Maclane functors. I gave a talk, of which this answer is a crude summary.

A personal comment. Nowadays, these kind of questions is completely out of fashion, because homotopy categories are not anymore the foundations of homotopical algebra. Not even their enhanced cousins derivators had much fortune after Joyal-Lurie's treatment of $$\infty$$-categories. After quite some juvenile reluctance, I myself stand on the side of $$\infty$$-categories, but I think it would be very interesting to investigate Freyd's conservative functor $$\text{HoTop} \to \text{Set}$$, as it indeed provides an interesting variant of homotopy groups that completely classifies spaces up to homotopy, and we really know nothing about its behavior.

• I read the link you provided right away (and upvoted the comment), it was interesting, thanks. But it doesn't seem to me that this gives any clarification regarding my question - I limited myself to finite complexes. So, I think any small category has a faithful functor in $\mathrm{Set}$: Yoneda embedding followed by taking the cartesian product of all the sets in the diagram. Apr 11, 2022 at 8:48
• Maybe it should be clarify that Ivan's answer adress the question of the existence of a complete algebraic invariant for the category of *all CW-complex". The OP was asking about finite CW-complex only, for which a (very impractical) complete algebraic invariant exists for trivial reasons because the category is small (take the Yoneda embedding for example...). Though it is not completely clear if one can be found that satisfies the finiteness requirement asked by the OP. Apr 11, 2022 at 13:59
• Sorry I am not familiar with all the category language but could you comment on where rational homotopy theory fits into this picture? It only works for nilpotent spaces but for those it gives a good algebraic model of rational homotopy equivalence (using either Sullivan's dga or Quillen's dgl constructions). it gives a conservative faithful functor from rationalizations of spaces and their homotopy equivalences to quasiisomorphism classes of of dgas. Does Freyd's theorem say something about possibility of similar functors for finer equivalences (between rational and full he)? Apr 11, 2022 at 14:44
• @VitaliKapovitch Homotopic morphisms between minimal Kan complexes need not be equal, even though homotopy equivalences must be isomorphisms. I don't understand minimal DGA's very well but I would expect it's the same issue here. In particular, I'm reasonably confident that Ivan and Fosco's theorem disproves concreteness of the rational homotopy category, though they do not prove this themselves. Apr 13, 2022 at 23:55
• Freyd's argument (and similar ones around this topic) all use a strong form of choice, in essence well-ordering the universe of sets. Forthcoming work of Karvonen and myself has removed Choice altogether, and even the need to use ZF as a foundation. Apr 14, 2022 at 10:30

I suspected that it would be possible to prove that there is no faithful functor from finite complexes to any kind of category of finite algebraic objects by slavishly following Freyd's argument and replacing "set" with "finite set" everywhere. This turned out to be true, and indeed the argument is a little bit easier in the group theory than the general case, so I reproduce it below as an advertisement for anybody lucky enough not yet to have followed Freyd's classic argument.

Proposition: There is no faithful functor from the homotopy category of finite complexes to the category of finite sets.

Proof: Consider the Moore spaces $$X_k=M(\mathbb Z_{2^k},2),$$ which are finite complexes. Let $$f_k:X_1\to X_k$$ be induced by the unique nontrivial homomorphism sending $$1$$ to $$2^{k-1}.$$ If $$F:\mathrm{Cell}^{\mathrm{op}}\to \mathrm{FinSet}$$ is a contravariant pointed functor into finite sets, then some $$F(f_k),F(f_j)$$ must have the same image, since $$F(X_1)$$ has only finitely many subsets. Without loss of generality, $$k\ge j.$$

Now observe that there is a map $$g:Y\to X_1$$ such that $$f_k\circ g=0$$ but $$f_j\circ g\ne 0.$$ Indeed, we can let $$f_k=\Sigma f_k'$$ and $$g$$ be the cone of $$f_k'$$. Then $$f_k$$ is a weak cokernel of $$g,$$ so $$f_k\circ g=0,$$ but if $$f_j\circ g$$ were $$0,$$ then we'd have a map $$h:X_k\to X_j$$ with $$h\circ f_k=f_j.$$ But this would imply the existence of a homomorphism $$\mathbb Z_{2^k}\to\mathbb Z_{2^j}$$ sending $$2^{k-1}$$ to $$2^{j-1},$$ which does not exist. Thus $$F(f_k\circ g)=0,$$ which means $$F(g):F(X_1)\to F(Y)$$ is zero on the image of $$F(f_k),$$ so that also $$F(f_j\circ g)=0.$$ Therefore $$F$$ is not faithful, and indeed, it sends a nonzero map to a zero map. $$\square$$

We have proved that there is no contravariant faithful functor from $$\mathrm{Cell}$$ to $$\mathrm{FinSet}.$$ Since $$\mathrm{FinSet}$$ admits a faithful contravariant endofunctor given by the powerset, there is also no covariant faithful functor $$\mathrm{Cell}\to\mathrm{FinSet},$$ so one might say that the homotopy category of finite complexes is not finitely concrete. Any reasonable category of finite algebraic objects will admit a faithful and conservative functor into $$\mathrm{FinSet},$$ so we can conclude that the answer to your question is negative.

Remark: Interestingly, I don't think you can carry this argument over to compact objects in Liberti-Loregian's generalization since they use $$\pi_n$$ instead of $$H_n,$$ and Eilenberg-Mac Lane spaces are less friendly to finiteness.

• Thanks for your answer, it was helpful for me! But the text of my question does not say that the resulting algebraic objects are finite. Archetypal examples of functors in the style of my question: homotopy groups, homology groups. Apr 15, 2022 at 9:48
• @AivazianArshak Yes, silly me, I read "finite set of carriers" as making the carrier(s) it(them)selves finite! Anyway you generously left some wiggle room in inviting answers to variants. We've got "all complexes are not concrete", "finite complexes are concrete", and "finite complexes are not finitely concrete" handled, now perhaps we can get to your actual question, "finite complexes do not embed fully faithfully in models of an finite essentially algebraic theory..." Apr 15, 2022 at 23:45