bounded generalized homology theory : isomorphism

Question

Let $h_*$ be a generalized homology theory. Write $X = \bigcup_{s \ge 0} X^s$, where $X^s$ is skeletal filtration of CW complex $X$.

It is claimed in Kochman's pg124, line 4 that:

if $P$ is the point and $h_k(P)=0$ for $k \le N$, then for any $s,t \in \Bbb Z$ $$j_*:h_{s+t}(X^{s+r}) \rightarrow h_{s+t}(X^{s+r+1})$$ is an isomorphism for sufficiently large $r$.

I'm wondering how the proof of this goes, and if there are any references for similar results. The result seems extremely similar to that of cellular cohomology.

Answer 1

Let $N \in \mathbb{Z}$ and let $h_*$ be a homology theory such that $h_n(P) = 0$ for all $n\leq N$ (for example, cellular/singular homology satisfies this for $N= -1$). What I managed to prove is that

for any $s<k$ the map $h_n(X^s) \to h_n(X^k)$ is an isomorphism for $n\leq s + N$.

This is equivalent to what you've stated since for a fixed homological degree it says you get an isomorphism for sufficiently large $s$ (in your case $k=s+1$). For a reference I'm not sure, but it may be in Spanier.

Consider the long exact sequence of the pair

$$\dots \to h_{n+1}(X^k,X^s) \to h_n(X^s) \to h_n(X^k) \to h_n(X^k,X^s) \to \dots $$

Then by exactness the map $h_n(X^s) \to h_n(X^k)$ is an isomorphism iff the homomorphisms on either side are both trivial, so it's sufficient that $h_{n+1}(X^k,X^s)\cong h_{n}(X^k,X^s) = 0$. Since $(X^k, X^s)$ is a "good pair" in the sense of Hatcher, for any $n$ with have the isomorphism

$$ h_n (X^k,X^s) \cong \tilde{h}_n(X^k/X^s)$$ (The proof given in Hatcher (Proposition 2.22) for singular homology still works here since it only uses homotopy invariance and excision.)

Consider first the case where $k-s=1$. Then $$ h_n (X^k,X^s) \cong \tilde{h}_n(X^k/X^s) \cong \tilde{h}_n\big(\bigvee S^k\big) \cong \bigoplus \tilde{h}_n(S^k)\cong \bigoplus \tilde{h}_{n-k}(S^0)\cong \bigoplus h_{n-k}(P)$$ So $h_{n+1}(X^k,X^s)\cong h_{n}(X^k,X^s) = 0$ if $n+1-k\leq N$. For a fixed dimension $k$ we see that we get an isomorphism in homology in degrees $n\leq N+k-1 = N +s$, or dually for a fixed homological degree $n$ we get a lower bound on the dimension of the complexes by $k\geq n-N +1$ (or $s \geq n-N$).

Now suppose $k-s >1$, and consider the chain of subcomplexes

$$X^s\subset X^{s+1} \subset \dots \subset X^{k} $$

If each of these maps induce an isomorphism in degree $n$ then their composition does as well. But for each $s\leq i < k$, the map $X^i \to X^{i+1}$ induces an isomorphism in homology in degree $n$ if $i\geq n-N$, therefore in fact any $k$ will work as long as $s\geq n-N$, or equivalently $n\leq s+ N$.