8
$\begingroup$

Let $h^*$ be a (multiplicative) generalized cohomology theory. Let $X$ be a CW complex which is a union of an increasing sequence $X_0 \subset X_1 \subset X_2 \subset \cdots$ of subcomplexes. Then there is a Milnor exact sequence $$0 \to (\varprojlim)^{1}(h^{n-1}(X_i)) \to h^n(X) \to \varprojlim h^n(X_i) \to 0.$$

Obviously, the restriction maps are ring maps and so the kernel $I=(\varprojlim)^{1}(h^{n-1}(X_i))$ gives an ideal.

Question: Is there anything that can be said about the multiplicative structure on the $(\varprojlim)^{1}(h^{n-1}(X_i))$ piece? In every example I find in textbooks, $I^2=0$, but I don't know if that's what happens in general.

$\endgroup$

1 Answer 1

12
$\begingroup$

Let $P$ be the wedge of all the $X_i$s. Up to homotopy equivalence, $X$ is the homotopy coequalizer of the identity and the shift maps from $P$ to itself. The Milnor exact sequence arises by analyzing the resulting long exact sequence $$ ...\rightarrow h^n(\Sigma P) \xrightarrow{1 - \text{shift}} h^n(\Sigma P) \xrightarrow{d^*} h^n(X) \xrightarrow{p^*} h^n(P) \xrightarrow{1- \text{shift}} h^n(P) \rightarrow \dots,$$ where $P \xrightarrow{p} X \xrightarrow{d} \Sigma P \xrightarrow{1-\text{shift}} \Sigma P$ is a cofibration sequence.

From this, one sees that an element $x \in h^n(X)$ is in $I$ (defined as you did) if and only if $x$ is in the image of $d^*: h^n(\Sigma P) \rightarrow h^n(X)$. Since cup products vanish in $h^*(\Sigma P)$ (the cohomology of a suspension), it follows that $I^2 = 0$, as you suspected.

[Homology groups here are all reduced.]

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.