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I'm in trouble in understanding these two statements in Morita's Geometry of Characteristic Classes book:

  1. First: enter image description here

    what's up with the "twisted" product $K(\pi_2(X),2)\times_{k^4} K(\pi_3(X),3)\cdots\times \cdots$? How is it defined?

  2. Second: enter image description here

    Is $\mathbb Q[\iota]$ the polynomial ring in the "variable"/cohomology class $\iota$? And what about $E_\mathbb Q(\iota)$: in which sense it is "the exterior algebra" over $\mathbb Q$?

I strongly suspect that the problem is "mine" in the sense that I'm not really into these topics. Any kind of reference is appreciated!

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2  
For (2): Yes $\mathbb{Q}[i]$ is the polynomial algebra generated by that class. And an exterior algebra makes sense over any field (with a bit of care at the prime 2). In this case it means something really easy: i^2 = 0, and that's it. – Dylan Wilson Dec 15 '12 at 23:17
2  
For (1): All this means is that given a $k$-invariant $k^{n+1}:X_{(n-1)} \to K(\pi_nX,n+1)$, you recover $X_{(n)}$ as the pullback of the path-loop fibration. This has fiber $K(\pi_nX,n)$, so it's a "twisted product" of $X_{(n)}$ with $K(\pi_nX,n)$ (i.e., a fibration). – Aaron Mazel-Gee Dec 16 '12 at 8:11
    
I think it's slightly better to think of the polynomial algebra and the exterior algebra as both being cases of a free dg commutative algebra on an even and odd degree generator, respectively. Then, the EM spaces give you the same kinds of algebras. – Justin Young Dec 17 '12 at 7:20

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