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For a path-connected space $X$, a covering space is a fiber bundle with a discrete set. It is known that if $X$ in addition locally path-connected and semilocally simply-connected, then $X$ has a universal cover.

Going over the construction of the universal cover, I see that it is the fibrant replacement of the inclusion of a point $x_0\hookrightarrow X$, i.e. replacing $x_0$ with the mapping path space $P(X,x_0)$ which is contractible and fibrates onto $X$.

I wonder what if we dropped the condition that $X$ is locally path-connected and semilocally simply-connected, then wouldn't the fibration $P(X,x_0)\to X$ still be very interesting? It will not be a universal cover per se, but will be a homotopy version of it in the sense of a fibration from a contractible space. What good properties still can be said for $P(X,x_0)$ and what are lost?

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    $\begingroup$ The universal covering space is not contractible in general. $\endgroup$ Feb 26, 2016 at 19:36
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    $\begingroup$ @mez: no, that's not correct. The universal cover has the same higher homotopy as $X$, so it is usually not contractible. $\endgroup$ Feb 26, 2016 at 19:48
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    $\begingroup$ You've missed that in the construction of the universal cover, paths from $x_0$ are identified if they're homotopical rel end points. This is in contrast to the construction of $P(X,x_0),$ where there's no such identification made. $\endgroup$ Feb 26, 2016 at 19:51

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The path space in the path space fibration is not the universal cover, even up to homotopy, unless $X$ is aspherical. Its fiber, rather than being the fundamental group, is the based loop space of $X$. But it is reasonable to think of it as a higher analogue of the universal cover.

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  • $\begingroup$ Okay, I see. So for the actual universal cover, we take end-points fixed homotopy classes of elements of mapping path space, that's what makes the difference right? $\endgroup$
    – mez
    Feb 26, 2016 at 19:50

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