# Product structure in Milnor exact sequence

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.

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.