Homology - long exact sequence from short exact sequences of cofibrations

639 Views Asked by At

I am currently self-studying a course in algebraic topology (http://ium.mccme.ru/f18/f18-topology-2.html, it's in Russian) and I've been stuck at one problem for quite some time.

Let $\mathcal{HoCW}$ be the category of homotopy-equivalent CW-complexes and classes of homotopic maps between them. We have a sequence of functors $h_n: \mathcal{HoCW} \rightarrow \mathcal{Ab}, n\in \mathbb{Z}$ such that

  1. There exist natural isomorphisms $h_n(X)\rightarrow h_{n+1}(\Sigma X)$.
  2. For any cofibration $A\to X$ in $\mathcal{HoCW}$ the sequence $h_n(A)\to h_n(X)\to h_n(X/A)$ is exact.
  3. For any bouqet $X=\bigvee_\alpha X_\alpha$ with inclusions $i_\alpha: X_\alpha\to X$ the direct sum $\bigoplus_\alpha (i_\alpha)_*:\bigoplus_\alpha h_n(X_\alpha)\to h_n(X)$ is an isomorphism.

The problem is to show the exactness of the sequence

$$\cdots\to h_n(A)\to h_n(X)\to h_n(X/A)\to h_{n-1}(A)\to\cdots$$ for any cofibration $A\to X$. Now, exactness of subsequences $h_n(A)\to h_n(X)\to h_n(X/A)$ comes directly from the axioms. Exactness in the terms $h_n(X)\to h_n(X/A)\to h_{n-1}(A)$ may be shown using the homotopy equivalence $X/A \simeq X \cup_A CA$ and the cofibration $X\to X\cup_A CA$. It remains to show the exactness of subsequence of the form $h_n(X/A)\to h_{n-1} (A)\to h_{n-1} (X)$, which I am struggling with.

1

There are 1 best solutions below

5
On BEST ANSWER

In fact, if $i : A \to X$ is a cofibration, then the quotient map $p : X \cup CA \to (X \cup CA)/CA = X/A$ is a homotopy equivalence. This shows that axiom 2 is equivalent to the exactness of $$h_n(A) \stackrel{i_*}{\rightarrow} h_n(X) \stackrel{j_*}{\rightarrow} h_n(X \cup CA)$$ where $j: X \to X \cup CA$ denotes inclusion. But now $j$ is again a cofibration and we get an exact sequence $$h_n(X) \stackrel{j_*}{\rightarrow} h_n(X \cup CA) \stackrel{k_*}{\rightarrow}h_n(X \cup CA \cup CX)$$ where $CX$ is glued to $X$ in the obviuos way. We could also write $X \cup CA \cup CX = CA \cup CX$, but let us leave it as it is. Proceeding this way we get a longer eaxct sequence $$h_n(X) \stackrel{j_*}{\rightarrow} h_n(X \cup CA) \stackrel{k_*}{\rightarrow} h_n(X \cup CA \cup CX) \stackrel{l_*}{\rightarrow} h_n(X \cup CA \cup CX \cup C(X \cup CA))$$ Since $CX$ is contractible, the quotient map $$q: X \cup CA \cup CX \to (X \cup CA \cup CX)/CX = CA/A = \Sigma A$$ is a homotopoy equivalence. Similarly $$r : X \cup CA \cup CX \cup C(X \cup CA) \to ((X \cup CA \cup CX \cup C(X \cup CA))/C(X \cup CA) \\= CX/X = \Sigma X$$ is one. Moreover it is easy to $r \circ l \simeq \Sigma i \circ q$. Now apply axiom 1.

You also see that axiom 3 is not needed.

Remark.

In your question you define a reduced homology theory for CW-complexes. In my opinion it is better to state the exactness axiom in the above form. Doing so we can work with more general spaces than CW-complexes and omit the assumption that $i : A \to X$ is a cofibration. The above arguments remain valid and we obtain a long eaxct sequence. See for example

Switzer, Robert M. Algebraic topology-homotopy and homology. Springer, 2017.