Product structure in Milnor exact sequence

Question

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.

Answer 1

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.]