This is definitely not a research level question. I believe this is "common sense" among homotopists, however after "extensive" googling for 2 days I could not find a proof of it online or in standard textbooks (Hatcher, Milnor-Stasheff, Husemoller). I asked on [mathstackexchange](https://math.stackexchange.com/questions/373005/naive-question-why-bmon-rightarrow-top-omega-top-rightarrow-mon-i) but did not receive an answer. I also tried to use online resources like nlab. It seems to be this is related to the bar construction, but I do not think the proof should be that complicated. I read in lecture notes that this is a natural consequence of the long exact sequence of fibration $F\rightarrow E\rightarrow B$, and the adjointness follows from homotopy lifting property, but I do not know how to put everything together. Below is the original post on Math.SE website. I also want to know "what should I read" for this kind of question, like why Bott periodicity can be framed as $\Omega U\cong \mathbb{Z}\times BU$.


It is not clear to me why we have a bijection of the form $$Mor\_{Top^{\*}}(BG,X)\rightarrow Mor\_{Mon}(G,\Omega X)$$where $Mon$ is the category of topological monoids and $Top^{\*}$ the based topological spaces. It seems to be something essentially trivial that the instructor did not bother to write down the proof in notes, but after thinking for 20minutes I still do not understand how to construct such a bijection to let $B$ and $\Omega$ be adjoint functors (or maybe I formulated it wrong somehow?). I think I need to this result to prove the well known result that $$\Omega BG\cong G, B\Omega X\cong X$$
Just to clarify definition, here $BG$ is the weakly contractible total space $EG$ quotient out by group action of $G$. $\Omega X$ is the group formed by mappings of the circle to $X$ with a fixed based point.


Sorry for the low level of this question.



Why $\Omega X$ and $BG$ are adjoint functors?
Are you willing to accept a proof of the adjunction in the homotopy category?


I think the more natural way to do that is to prove first that $\Omega B G \simeq G$ (homotopy equivalence) and $B \Omega X \simeq X$ (note that $X$ needs to be connected for this second statement to be true, so you have to restrict your category). The first equivalence follows from writing out the long exact sequence in homotopy for the fiber sequence $G \rightarrow EG \rightarrow BG$. (You use Whitehead's theorem to pass from an isomorphism on all homotopy groups to a homotopy equivalence.) The second equivalence follows from comparing the fiber sequence $\Omega X \rightarrow P X \rightarrow X$ to the fiber sequence $\Omega X \rightarrow E \Omega X \rightarrow B \Omega X$.


Then if you have a map $BG \rightarrow X$, you loop it and get (up to homotopy) a map $G \rightarrow \Omega X$, and if you have a topological monoid morphism $G \rightarrow \Omega X$, you apply the bar construction (which is functorial) to get (up to homotopy) a map $B G \rightarrow X$. These are inverses on homotopy classes.


