TITLE: Does there exist a complete algebraic invariant of the homotopy type of a finite CW-complex?
QUESTION [7 upvotes]: 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.

REPLY [6 votes]: 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.