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.
 A: 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.
A: 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.
