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Tom Goodwillie
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I think that homotopy-theorists often fall into the habit of working mainly with based spaces, even when they don't need to. It can be instructive to notice when the use of a basepoint is unnecessary, even artificial. But it's also important to notice the parts of the subject where the use of a basepoint is necessary. This (the topic of universal covering spaces) is one of those parts.

By "universal covering space" of a connected manifold $M$ I assume we mean a simply connected manifold $\tilde M$ with covering map $p:\tilde M\to M$. (By "simply connected" I mean connected and having trivial $\pi_1$ for one, hence any, basepoint. The empty space is not connected.)

There is always a universal covering space, and to explain how to make one we usually start by picking a point $x\in M$. Any two universal covering spaces, no matter how they are constructed, are related by an isomorphism, by which I mean a diffeomorphism that respects the projection to $M$. But this isomorphism is not unique, because for any such $(\tilde M,p)$ there is a group of such isomorphisms $\tilde M\to \tilde M$ (i.e. deck transformations), a nontrivial group except in the case when $M$ itself is simply connected.

No matter how we interpret the question being asked here, I think we have to say that an affirmative answer would imply that the isomorphism between two universal covering spaces of $M$ can be made canonical -- independent of any choices.

But now look at the case where two universal covering spaces of $M$ are both made in the usual way, by choosing a point in $M$. Say that we make one covering space using $x\in M$ and the other using $y\in M$. Every homotopy class of paths from $x$ to $y$ in $M$ (homotopy with endpoints fixed) determines an isomorphism between them, and every isomorphism arises from exactly one such homotopy class. So if we had a canonical isomorphism we would have a canonical homotopy class of paths from $x$ to $y$. And surely we don't.

Tom Goodwillie
  • 55.9k
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  • 240