Let $h_n\colon CW_*\to Ab (n\in \mathbb{Z})$ be a covariant homotopy invariant functor that sends cofibre sequences to exact sequences and is equipped with natural suspension isomorphisms.

Then the following two statements are equivalent:

  1. For every countable collection $(X_i)_{i\in I}\subseteq CW_*$ the natural morphism $\bigoplus_i h_*(X_i)\to h_*(\bigvee_i X_i)$ is an isomorphism.

  2. If $Y=colim_k(Y_0\overset{cl.incl.}{\hookrightarrow}Y_1\hookrightarrow\cdots)$, then the natural morphism $colim_k h_*(Y_k)\to h_*(Y)$ is an Isomorphism.

I tried to show this, and had some ideas, but I still lack of understanding for a detailed proof. In particular, it is not clear to me how the suspension isomorphisms and the cofibre condition have to be used.

My considerations are as follows:

$2.\implies 1.$ Assume 2. is true. Let $(X_i)_i$ be any countable collection of pointed CW-complexes. We may assume, that the index set are the natural numbers. The most obvious thing that comes to my mind is to let $Y_k:=X_0\vee \cdots \vee X_k$. Then let $Y=colim(Y_0\hookrightarrow Y_1\hookrightarrow \cdots)$.

(one issue here is, that I don't know how to replace the inclusions by closed ones, so I neglect this for a moment)

Now by 2. we know that the natural morphism $colim_k h_*(Y_k)\to h_*(Y)$ is an isomorphism.

My intuition tells me that $h_*(X_0\vee\cdots \vee X_k)\cong h_*(X_0)\oplus \cdots \oplus h_*(X_k)$ should be true under the above assumptions, and that one has $colim_k (h_*(X_0)\oplus\cdots\oplus h*(X_k))\cong \bigoplus_i h_*(X_i)$.

$1.\implies 2.$ Assume 1. is true. Let $Y=colim(Y_0\hookrightarrow Y_1\hookrightarrow\cdots)$, where the $hookrightarrow$ are closed inclusions. If one could make the $Y_k$ become disjoint, the relations between them would disappear and the colimit would become a coproduct. Hence one could replace $Y$ by the reduced mapping telescope $\bigcup_{k\geq 0} [k,k+1]\times Y_k$ and show that this doesn't affect the value of $h_*$. By one we would have $\bigoplus_k h_*(Y_k)\cong colim_k h_*([k,k+1]\times Y_k)\cong colim_k h_*(Y_k)$.

-Is there a way to make this argument work?

-How do the suspension isomorphisms and the cofibre condition join the argument?

  • $\begingroup$ You consider a reduced homology theory on $CW_*$. It is well-known that then $h_*(X \vee Y) \approx h_*(X) \oplus h_*(Y)$ which generalizes to finite wedges. Moreover, the inclusion $X_1 \vee \dots \vee X_k \hookrightarrow X_1 \vee \dots \vee X_{k+1}$ is closed. $\endgroup$ – Paul Frost Jan 7 at 9:29
  • $\begingroup$ 1. $\Rightarrow$ 2. is not all trivial. You can find a proof in Switzer, Robert M. Algebraic Topology - Homotopy and Homology. Springer, 2017. See Proposition 7.53 which has a two-pages-proof (plus the material proved previously). $\endgroup$ – Paul Frost Jan 7 at 13:28

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