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Suppose I have an $A_{\infty}$-space $X$, such that its unit is only a unit up to homotopy. When the space is well-behaved (well-pointed? What is the weakest condition possible?), I can replace it with a homotopy equivalent version of $X$ that has an honest unit. I read the definition for the classifying space of an $A_{\infty}$-space in Stasheff's papers. He uses honest units. Is it possible to circumvent this somehow?

Is there a functorial definition of the classifying space $BX$ that does not strictify the homotopy unit?

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1 Answer 1

up vote 9 down vote accepted

For your first question:

If $X$ has the homotopy type of a CW space, then you can replace $X$ by any CW space $Y$ that is homotopy equivalent to it (in the unbased sense).

Then $Y$ is also $A_\infty$ with a good basepoint. Then you can use the homotopy extension property to make the basepoint a strict unit for the multiplication.

As to your last question:

I think the answer is yes. For example, one can use the Boardman and Vogt description of the classifying space; this is a model which doesn't require that the unit be strict.

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Boardman and Vogt use what they call WA-spaces and they write in Homotopy-Everything H-spaces: "A WA-space (...) is approximately an $A_{\infty}$-space." I don't understand the "approximately"-part. How are these two notions related? –  Ulrich Pennig Nov 8 '11 at 23:01
    
Ulrich, I recommend that you go instead to the introduction of their book--it's explained there. –  John Klein Nov 8 '11 at 23:29
    
The book seems to be exactly what I was looking for. Thank you! –  Ulrich Pennig Nov 9 '11 at 7:18

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