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 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.

(That's not rigorous, because what does "canonical" mean? But surely if one had an actual recipe for making an $\tilde M$ for $M$ without first making some arbitrary choice then for any diffeomorphism $h:M_1\cong M_2$ the choice of canonical path classes in $M_1$ would be related by $h$ to the corresponding choice in $M_2$. In particular this would be the case for a reflection $S^1\to S^1$ that fixes two points $x$ and $y$ but of course does not fix any class of paths from $x$ to $y$.)

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 isomorphisms $\tilde M\to \tilde M$ (i.e. deck transformations), a nontrivial group except in the case when $M$ itself is simply connected.

Suppose that there were a way of making a universal covering space $\tilde M$ that did not depend on a choice of basepoint (or any other arbitrary choice), and suppose that for $x\in M$ there was a canonical isomorphism between this $\tilde M$ and the one determined by $x$.

But this would imply that when we use two points $x\in M$ to make two universal covering spaces of $M$ then there is a canonical isomorphism between these. 

Every homotopy class of paths from $x$ to $y$ in $M$ (homotopy with endpoints fixed) determines an isomorphism between the two covering spaces, 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.

(That's not rigorous, because what does "canonical" mean? But surely if one had an actual recipe for making an $\tilde M$ for $M$ without first making some arbitrary choice then for any diffeomorphism $h:M_1\cong M_2$ the choice of canonical path classes in $M_1$ would be related by $h$ to the corresponding choice in $M_2$. In particular this would be the case for a reflection $S^1\to S^1$ that fixes two points $x$ and $y$ but of course does not fix any class of paths from $x$ to $y$.)

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.

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 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.

(That's not rigorous, because what does "canonical" mean? But surely if one had an actual recipe for making an $\tilde M$ for $M$ without first making some arbitrary choice then for any diffeomorphism $h:M_1\cong M_2$ the choice of canonical path classes in $M_1$ would be related by $h$ to the corresponding choice in $M_2$. In particular this would be the case for a reflection $S^1\to S^1$ that fixes two points $x$ and $y$ but of course does not fix any class of paths from $x$ to $y$.)