Homology Theory – Satisfying Milnor’s Additivity Axiom but Not Direct Limit Axiom

at.algebraic-topologyhomology

Let us agree on the following: a "homology theory" means a functor $h_*$ from the category of pointed CW complexes to the category of graded abelian groups, together with natural isomorphisms $h_{*+1}(\Sigma X)\cong h_*(X)$, such that the functor $h_*$ is homotopy invariant, sends a cofiber sequence to an exact sequence, and satisfies Milnor's additivity axiom: the natural map to the homology of a wedge from the direct sum of the homologies is an isomorphism.

Adding Milnor's axiom to the list is a reasonable thing to do, partly because it guarantees uniqueness in the following sence: a morphism of homology theories (satisfying Milnor's axiom) which induces an isomorphism on the coefficients, is itself an isomorphism of homology theories.

Now enter the world of spectra. A spectrum $E$ defines a homology theory via $E_*(X)=\pi_*(E\wedge X)$. Such a homology theory satisfies an axiom a priori stronger than Milnor's additivity: it satisfies the

Direct limit axiom: The natural map to the homology of $X$ from the colimit of the homologies of the finite subcomplexes of $X$ is an isomorphism.

Reciprocally, a homology theory satisfying this axiom comes from a spectrum in the way defined above (see Adams' Stable Homotopy and Generalised Homology, pp. 199-200 and Adams "A Variant of E.H. Brown's representability theorem".)

So the homology theories given by spectra are a bit better behaved than a homology theory as is usually defined. The question is:

Could you give an example of a homology theory which does not satisfy the direct limit axiom?

I first asked this question on a comment here, and Mark Grant pointed out that by Hatcher's exercise 4.F.1, Milnor's additivity implies the direct limit axiom for countable complexes, so we'd need to go to uncountable ones.

Best Answer

I think it's a great question because not everybody is aware of this. A consequence of Brown representability theorem (Adams's version, which is highly non-trivial and depends on homotopy groups of spheres being countable) is that any spectrum $X$ fits into an exact triangle \[\coprod_{j\in J}Z_j\longrightarrow \coprod_{i\in I}Y_i\longrightarrow X\stackrel{\delta}\longrightarrow\Sigma Z\]

where $\delta$ is a phantom map and $Y_i$ and $Z_j$ are finite CW-spectra for all $i\in I$ and $j\in J$, see Neeman's "On a theorem of Brown and Adams". Hence taking homology we obtain a short exact sequence

$$\coprod_{j\in J}h_{*}(Z_j)\hookrightarrow \coprod_{i\in I}h_{*}(Y_i)\stackrel{p}\twoheadrightarrow h_{*}(X).$$

You can (and do) take without loss of generality $\{Y_i\}_{i\in I}$ to be the family of finite subcomplexes of $X$.

We also have by definition an exact sequence (not injective on the left)

$$\coprod_{\{Y_k\subset Y_i\}}h_{*}(Y_k)\longrightarrow \coprod_{i\in I}h_{*}(Y_i)\stackrel{q}\twoheadrightarrow \operatorname{colim}_{i\in I}h_{*}(Y_i).$$ The previous surjection $p$ factors throug $q$ and the canonical map $\operatorname{colim}_{i\in I}h_{*}(Y_i)\rightarrow h_*(X)$.

The first map is induced by obvious vertical map in the diagram below, and the diagonal factorization exists by general properties of triangulated categories $$ \begin{array}{ccccc} &&\coprod_{\{Y_k\subset Y_i\}}Y_k&&\\ &\swarrow&\downarrow&&\\ \coprod_{j\in J}Z_j&\longrightarrow &\coprod_{i\in I}Y_i\longrightarrow& X \end{array} $$ It is only left to show that the diagonal map is surjective on homology. This map actually has a splitting in the stable homotopy category (see Neeman's paper), hence we're done.

All this is very specific to the stable homotopy category, Adams's theorem is seldom satisfied elsewhere, hence model category arguments would usual fail in proving this.