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.


3 Answers 3


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.


If we assume the additivity axiom for all coproducts (not just countable ones), then a homology theory carries all filtered homotopy colimits to colimits of abelian groups.

Let $J$ be a filtered category and given a homology theory $h_*$, consider two families of functors $\mathsf{Top}^J \to \mathsf{Ab}$. One is $X \mapsto h_*(\mathrm{hocolim}_{j \in J} X_j)$ and the other is $X \mapsto \mathrm{colim}_{j \in J} h_* X_j$. Both these families of functors are "homology theories" in the sense that they satisfy all the same axioms as $h_*$ (using the fact that $J$ is filtered to show that the latter is exact).

The universal properties of (homotopy) colimits yield a transformation which is an isomorphism "on coefficients" which in this case means on every representable functor over $J$. It follows (by the usual argument you mention in your question) that this transformation is an isomorphism everywhere. In this case, you need to observe that every diagram of spaces over $J$ is equivalent to a projectively cofibrant diagram, i.e. one obtained by pushing out coproducts of $J(j, -) \times S^n \to J(j, -) \times D^{n+1}$ and then taking countable sequential colimits. Compactness of $S^n$ implies that countable colimits in the small object argument suffice. Now we proceed by induction using exactness to handle the pushouts, additivity to handle the coproducts and Hatcher's exercise to handle the colimit of a countable sequence.

Added: the argument of the previous paragraph in more detail. Given diagrams $A, B \colon J \to \mathsf{Top}$, a map $f \colon A \to B$, $j \in J$, $n \in \mathbb{N} \cup \{ -1 \}$, consider the set $C_{j,n} (f)$ of all commutative squares$\require{AMScd}$

\begin{CD} J(j, -) \times S^n @>>> A \\ @VVV @VV{f}V \\ J(j, -) \times D^{n+1} @>>> B. \end{CD}

Define $\bar B$ by pushing out $\bigsqcup_{j,n} C_{j,n} (f) \times J(j, -) \times S^n \to \bigsqcup_{j,n} C_{j,n} (f) \times J(j, -) \times D^{n+1}$ along $\bigsqcup_{j,n} C_{j,n} (f) \times J(j, -) \times S^n \to A$.

Now, given any diagram $X \in \mathsf{Top}^J$, define $X^{(-1)} = \varnothing$ and $f_{-1} \colon X^{(-1)} \to X$ as the unique map. Then apply the above construction to get $X^{(0)} = \bar X^{(-1)}$ and the induced map $f_0 \colon X^{(0)} \to X$. Proceed inductively to obtain a sequence of maps $X^{(-1)} \to X^{(0)} \to X^{(1)} \to \ldots$ and let $\tilde X$ denote its colimit which comes with an induced map $\tilde X \to X$. This latter map is a weak equivalence, even an object-wise acyclic Serre fibration. Indeed, in any a square

\begin{CD} S^n @>>> \tilde X_j \\ @VVV @VVV \\ D^{n+1} @>>> X_j \end{CD}

the top map factors through some $X_j^{(k)}$ since $S^n$ is compact and the resulting square has a lift since it corresponds by an adjunction to one of the squares above and a solution to that lifting problem was adjoined in the construction of $X^{(k+1)}$.

Now, assume that $H_*, K_* \colon \mathsf{Top}^J \to \mathsf{Ab}$ are homology theories with a transformation $\phi \colon H_* \to K_*$ that is an isomorphism on representables. We proceed to prove that it is an isomorphism everywhere in a few steps.

Step 1. $\phi$ is an isomorphism on $J(j, -) \times D^{n+1}$ by homotopy invariance.

Step 2. $\phi$ is an isomorphism on $J(j, -) \times S^n$ by the usual Mayer--Vietoris argument (decompose the sphere into two hemispheres).

Step 3. $\phi$ is an isomorphism on both $\bigsqcup_{j,n} C_{j,n} (f_k) \times J(j, -) \times S^n$ and $\bigsqcup_{j,n} C_{j,n} (f_k) \times J(j, -) \times D^{n+1}$ by additivity.

Step 4. $\phi$ is an isomorphism on all $X^{(k)}$ by induction on $k$ applying exactness each time.

Step 5. $\phi$ is an isomorphism on $\tilde X$ by the mapping telescope trick.

Step 6. $\phi$ is an isomorphism on $X$ by homotopy invariance.

  • $\begingroup$ I don't have enough familiarity with some of the objects you use in your answer to understand it just yet. But if I understand correctly, you're actually saying that Milnor's additivity axiom is equivalent to the direct limit axiom, correct? $\endgroup$ Jan 29, 2016 at 14:36
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    $\begingroup$ Yes, my statement is rather abstract, but any CW-complex is the homotopy colimit of its finite subcomplexes so the "direct limit axiom" follows. If you have any specific questions, I can try to elaborate. $\endgroup$ Jan 29, 2016 at 14:40
  • $\begingroup$ Well, that part I understand. I'm still puzzled by the fact that additivity would imply taking all filtered homotopy colimits to colimits. Why would Adams have introduced the axiom in his 1970 paper then? Also, more recent books like Hatcher also insist on the necessity of this "stronger" form (p. 455). (What I don't understand is the more model-category-heavy part. But there's too much knowledge I still lack in that department, so don't worry trying to expand on it too much. Thanks, though.) $\endgroup$ Jan 29, 2016 at 14:47
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    $\begingroup$ Well, in the times of Adams things like that were not sorted out yet. Why Hatcher says that this axiom is stronger (or if he actually claims that it is strictly stronger), I cannot say. But in the recent years, abstract homotopy theory (including Goodwillie calculus which puts homology theories in an abstract context) progressed to a point that things like that became folklore. $\endgroup$ Jan 29, 2016 at 15:04
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    $\begingroup$ Actually, now that I look more closely, if we look at Switzer, Remark 1 p. 331, he says: "One can show that (the wedge axiom) implies (the direct limit axiom): the proof is a generalization of (the fact that homology commutes with colimits over a countable tower) using a "generalized telescope"", so, a homotopy colimit. I hadn't seen this before... $\endgroup$ Jan 29, 2016 at 15:19

Edit: As explained in the comments, this is not an answer to the question. I leave it as it might at least serve as an intuition for why the direct sum axiom plus excision should be enough to imply the direct limit axiom.

Working $\infty$-categorically, we may think of a homology theory as a functor from spaces to spectra. The axioms about the suspension isomorphism and cofiber sequences are equivalent to preservation of pushouts. Now, a functor which preserves pushouts and all (small) coproducts, preserves all (small) colimits. This is proposition in HTT by Lurie.

Remark: Actually, we can reverse the logic by saying that we just want our the functor to preserve all (small) colimits. One can break this into (1) preservation of finite colimits and (2) preservation of filtered colimits. One can also replace (2) by (2') preservation of coproducts. Additionally, one can replace (1) by (1') preservation of pushouts. Finally, one can test that a map of spectra is an equivalence by checking the induced map on the homotopy groups and for a pushout these can be described by a long exact sequence. This allows us to reformulate everything in classical terms (i.e. the Eilenberg-Steenrod axioms).

  • $\begingroup$ I think this is misunderstanding the question: the OP is interested in whether a functor to abelian groups that satisfies excision and the direct sum axiom is represented by a spectrum. As you notice, the question for functors to spectra is much easier. $\endgroup$ May 18, 2017 at 18:31
  • $\begingroup$ You are right. I read the question again and now I see that I misinterpreted it somehow. $\endgroup$
    – KotelKanim
    May 18, 2017 at 20:55

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