# Why is a path-connected topological space homotopy equivalent to the classifying space of its loop space?

Given a path-connected topological space $X$ (lets say compactly generated; this entire post will be working in the category of compactly generated topological spaces) with a designated point $x$, we can form the loop space $\Omega X$ of pointed morphisms from $S^1$ to $X$. This has the natural structure of a topological group.

Now given a topological group $G$, we can form its classifying space $BG$. I believe that it is true that $X$ is homotopy equivalent to $B\Omega X$; is this true?

I suspect that you could form some model category of topological groups and some model category of connected topological spaces, and show that $\Omega$ and $B$ are a Quillen equivalence. For one thing, I know that $\Omega BG$ is homotopy equivalent to $G$, and I have a feeling that if I could write out the model structures this should become clear. On the other hand I'm not that fluent in model category theory, and I'm also wondering if there's an elementary way to see this.

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When you say "homotopic," you mean "homotopy equivalent." –  Grumpy Parsnip Mar 21 '13 at 3:00
Errr, yes, sorry about that. –  Nehsb Mar 21 '13 at 3:03
Actually, what is the group structure on $\Omega X$? There are no inverses, and the multiplication is not associative (just homotopy-associative.) –  Grumpy Parsnip Mar 21 '13 at 3:12
You might look at this question and discussion for inspiration: mathoverflow.net/questions/41525/… –  Grumpy Parsnip Mar 21 '13 at 3:12
Oh, I was working with it having the structure of the group under homotopy (in which case reversing the path is an inverse) but I also thought it was an actual group (which it clearly is not.) In this case, there's still a classifying space (from Segal's paper "Categories and Cohomology Theories") but I didn't want to cite that... Also the linked thread is very helpful, thanks! (Especially the comment of Tom Goodwillie which states that there are multiple possible deloopings, which shows that (perhaps some variant) of my question is false, as my question implies all deloopings are equivalent. –  Nehsb Mar 21 '13 at 10:45