-2
$\begingroup$

Let the topological monoid $M$ be the configuration space $C(\mathbb{R}^n;X)=C_n(X)$ as in the book The geometry of iterated loop spaces, Theorem 5.2. I want to prove that the map $\alpha_n$ in Theorem~5.2 there is the same map as the one described in the answer to this question on math.stackexchange. How to prove?

$\endgroup$

1 Answer 1

5
$\begingroup$

$C_n(X)$ is not a monoid in any natural way; I never said it was. And the target $\Omega^n\Sigma^n X$ of $\alpha_n$ has $n$ different loop products. The question is not meaningful as posed. Nevertheless the proof of Theorem 6.1 (the approximation theorem) answers it by describing $\alpha_n$ as the fiber map of a map from a quasifibration to "the" path loop fibration $\Omega \Omega^{n-1}\Sigma^n X\to P \Omega^{n-1}\Sigma^n X \to \Omega^{n-1}\Sigma^n X$ (which one being determined by an ordering of the loop coordinates). From here, there are papers by Fiedorowicz and by Thomason that compare this $1$-fold delooping to the classifying space delooping of monoids given by Moore loop spaces.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.