The statement in the title seems to be generally accepted as true, but I have not seen proof. They are?


The strict formulation I have in mind is the following. By an algebraic category we mean the category of algebras of some monad on $\mathrm{Set}^S$. In particular, this includes all the usual finitary algebraic categories: groups, modules, etc [1]. Also note that the product of every set of algebraic categories is an algebraic category (this way we can combine every set of invariant into one).


**Question**. Is there an essentially injective functor $F: \mathrm{Hmpt} \to A \times B^{\mathrm{op}}$, where $\mathrm{Hmpt}$ is the category of homotopy types (which can be equivalently defined as the homotopy category of cw-complexes, delta-generated spaces, simplicial sets, or infinity-groupoids), and $A, B$ are algebraic categories (so we can use both covariant and contravariant invariant). Essentially injectivity means that $F(X) \cong F(Y)$ implies that $X \cong Y$.


For example, $F$ can take all of the following data


* all homotopy groups with their Whitehead product (in particular, the action of $\pi\_1$ on higher homotopy groups is stored)
* some set generalized homology and cohomology theories (we can't take all of them because they form a proper class) with all cohomological operations


**Sub-question**. Is this $F$ essentially injective?


The closest thing I can think of is that by [Freyd's theorem](http://www.tac.mta.ca/tac/reprints/articles/6/tr6.pdf) there can't be such a faithful functor, because $A \times B^{\mathrm{op}}$ is [concrete](https://ncatlab.org/nlab/show/concrete+category) over $\mathrm{Set}$. But I don't see how that would help. At the same time, it seems to me really curious to know whether there can be some set of algebraic invariants, by comparing which, we can conclude that the spaces are homotopically equivalent.


[1]: In addition to them, we have categories where operations in algebras can have infinite arity, as well as categories where the arity of operations is unlimited - the latter gives, for example, compact Hausdorff spaces. This doesn't bother me at all, because firstly these categories still behave quite similarly to algebraic ones (and seem to still be quite efficient invariants), and secondly, I think you can replace $A \times B^{\mathrm{op}}$ to any concrete category and the answer should still be no.



Can we prove that any number of algebraic data cannot be a complete invariant of a homotopy type?
I'll answer the corresponding question for the homotopy category $\mathcal{S}$ of spectra. I doubt that this makes much difference, but I have not checked the details. We can choose a list $X\_0,X\_1,\dotsc$ containing one representative of every homotopy equivalence class of finite spectra, and then put $X=\bigvee\_iX\_i$ and $R=[X,X]$. This has an idempotent $e\_i$ for each $i\in\mathbb{N}$, corresponding to the obvious map $X\to X\_i\to X$. Define $F\colon\mathcal{S}\to\text{Mod}\_R$ by $F(E)=E\_0R=\pi\_0(E\wedge R)$. Given $F(E)$ we can use the idempotents $e\_i$ to determine the groups $E\_0X\_i$, and $R$ contains $e\_jRe\_i=[X\_i,X\_j]$ so we can determine the full homology theory on finite spectra represented by $E$. It is a standard consequence of Brown representability theory that this determines $E$ up to a weak equivalence (which is itself unique up to phantoms).


Of course there is no hope of understanding this $R$ explicitly, so this answer may be considered unsatisfying, but it does answer the question as asked.


Here are some similar questions where positive answers are known.


* The category of rational spectra is equivalent to the category of graded rational vector spaces. More generally, let $G$ be a finite group, let $H\_1,\dotsc,H\_r$ be a list containing one subgroup from each conjugacy class, and let $W\_i=N\_G(H\_i)/H\_i$ be the associated Weyl group, and put $R=\prod\_i\mathbb{Q}[W\_i]$. Then the category of rational $G$-spectra is equivalent to the category of graded $R$-modules.
* Fix a prime $p$, and let $\mathcal{F}\_p$ be the category of spectra that can be expressed as the $p$-completion of a finite spectrum. We then have a functor $\pi\_\*\colon\mathcal{F}\_p\to\text{Mod}\_{\pi\_\*(S\_p)}$. Freyd conjectured (the ``Generating Hypothesis'' or GH) that this is faithful. It is known that many consequences would follow if GH were true; in particular $\pi\_\*(X)$ would be an injective module over $\pi\_\*(S\_p)$ for all $X\in\mathcal{F}\_p$, and the functor $\pi\_\*\colon\mathcal{F}\_p\to\text{Mod}\_{\pi\_\*(S\_p)}$ would in fact be a full and faithful embedding, so $\mathcal{F}\_p$ would be an algebraic category.
* Let $\mathcal{M}$ be the category of Moore spectra, i.e. spectra $X$ with $\pi\_i(X)=0$ for $i<0$ and $H\_i(X)=0$ for $i\neq 0$. Define a *Moore diagram* to be a diagram $A\xrightarrow{\phi}B\xrightarrow{\psi}A$ with $\psi\phi=0$ and $\phi\psi=2.1\_B$. We say that such a diagram is *exact* if the induced sequence $A/2\to B\to\text{ann}(2,A)$ is short exact. It can be shown (<https://arxiv.org/abs/1205.2247>) that the category of Moore spectra is equivalent to the category of exact Moore diagrams. However, even though this is a rather simple example, there is a surprisingly large amount of work in the proof, which is not a promising sign if you want to generalise further.
* You could relax your requirements and ask for good functors from a category of spectra to the derived category of an abelian category, rather than to the abelian category itself. The most important example is as follows. We can fix a prime $p$ and an integer $n>0$ and consider the category $\mathcal{L}(p,n)$ of spectra that are Bousfield-local with respect to the Johnson-Wilson spectrum $E(p,n)$. This category is a central player in the chromatic approach to stable homotopy theory. It is known that when $p$ is large relative to $n$, the category $\mathcal{L}(p,n)$ is equivalent to a kind of derived category of differential graded comodules. The first results in this direction were due to Franke, but the paper <https://arxiv.org/abs/1903.10003> (by Barthel, Schlank and Stapleton) is probably the best place to look to understand the current status.


