Hi,
Up until now I've kind of taken this issue for granted, but I'm trying to think about it now and nothing seems to be making the situation much clearer.
The problem is (in short), what the correct notion of the based loop space (or even based path space)should be for 1-connected spaces $X$ with a free $G$-action ($G$ a discrete group). I am requiring the freeness condition because what I am really interested in is the quotient space, and more precisely the action of the fundamental group ($G$) on the higher homotopy, and so the freeness condition is to ensure that $X$ is the universal cover of $X/G$. The way you would like to define an action (simply by using the action on X to induce an action on $Maps(S^1, X)$) doesn't work straight off, as it's not well defined: since the action is free, for any choice of basepoint, the constant loop will be moved by any non-identity element to a loop not based at the correct basepoint.
Morally, because equivariantly points are supposed to correspond to orbits in some way, I feel like the answer (for the equivariant path space $PX$) should be something like $PX=Maps(EG\times [0,1], X)$, where $[0,1]$ has the trivial G-action and a basepoint $0$, and we use a free $G$-cell construction of $EG$ and $X$ with exactly one equivariant $0$-cell in each, and the maps in the $PX$ are required to fix this $0$-cell (on the $[0,1]$ side, the $0$-cell appears as $G\times {0} $), and of course, all maps are $G$-maps. This still doesn't feel ideal, as we don't have a specified basepoint, and I don't know whether (in the usual $G$-model structure) the homotopy pullback of $PX\to X\leftarrow PX$ is the loopspace, which (under the above story) would be $\Omega X=Maps(EG\times S^1, X)$.
I guess more generally my question is how one defines an action on homotopy pullbacks in the equivariant case? I know there must be an answer, as the G-model structure does exist with fibrations and weak equivalences those which are fibrations and weak equivalences on fixed point spaces. Any help or comments would be much appreciated.
Thanks,
Tom