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Mar 9 at 20:15 comment added Tyrone @ThorbenK the rational $A_\infty$-equivalence will push down to the classifying space.
Mar 9 at 19:19 comment added LSpice @RyanBudney, re, thanks! I see the author got it.
Mar 9 at 18:37 comment added ThorbenK @Tyrone but does it suffice that the group has a free graded commutative cohomology ring over the rationals? Does this push down to the classifying space?
Mar 9 at 11:38 comment added Tyrone It will be true of any topological group (or even space) whose rational cohomology is a free associative, graded-commutative algebra over $\mathbb{Q}$. Simply choose a set of primitive generators to obtain a rational equivalence with a product of EM spaces. It's easy to verify that the obstructions for this to be an $A_\infty$-equivalence vanish. On the other hand, these cohomological conditions clearly need to hold for the outcome to be true. They're satisfied, for instance, by any group homotopy equivalent to a finite CW complex (e.g. any Lie group, as you recognise).
Mar 9 at 7:55 history edited ThorbenK CC BY-SA 4.0
edited title
Mar 9 at 6:46 comment added Ryan Budney @LSpice: you missed the title.
Mar 8 at 21:31 comment added Dave Benson I believe it was for the sake of his first wife, Dorothy Jones, who found it easier to type it with a space.
Mar 8 at 21:20 comment added LSpice Someone explained to me, but I can't remember the details of, the orthography of Mac Lane's name; he latterly went by Mac Lane, but originally by MacLane. (Maybe that's all there is to the explanation, but I seem to remember more.) But it was never Maclane, so I edited accordingly.
Mar 8 at 21:19 history edited LSpice CC BY-SA 4.0
Maclane -> MacLane
Mar 8 at 20:21 history edited ThorbenK CC BY-SA 4.0
added 2 characters in body
Mar 8 at 19:53 history edited ThorbenK CC BY-SA 4.0
added 84 characters in body
Mar 8 at 19:36 history asked ThorbenK CC BY-SA 4.0