Let $X$ be a real manifold (for simplicity). The standard construction of the universal cover $\varphi: \widetilde{X} \longrightarrow X$ involves fixing a basepoint $p \in X$ and considering homotopy classes of paths from $p$ to $x \in X$.
Is there an alternative construction of $\varphi$ that avoids choosing a basepoint?
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 $y$.)
Part of 10.5.8 of Topology and Groupoids is, in a more usual notation, essentially the following, in which $\sigma, \tau$ are the source and target maps, $St_G x$ is $\sigma ^{-1} x$, by $N$ is totally disconnected is meant that $N(x,y)$ is empty for $x \ne y$, and normality of $N$ in $G$ also means that $N,G$ have the same set of objects:
Let $X$ be a space which admits a universal cover, and let $N$ be a totally disconnected normal subgroupoid of the fundamental groupoid $\pi_1( X) $, Then the set of elements of the quotient groupoid $\pi_1(X)/N$ may be given a topology such that the projection $$q = (\sigma, \tau) : \pi_1(X)/N \to X \times X$$ is a covering map and for $x \in X$ the target map $\tau :St_{\pi_1( X)/N} \to X$ is the covering map determined by the normal subgroup $N(x)$ of $\pi_1(X, x)$.
So this uses all the points of $X$ and puts all these covers into a covering space, which means you don't make a choice of base point; instead you use all the choices. Further, $\pi_1(X)/N$ with this topology is actually a topological groupoid.
This may be the optimal way of answering the question.
I believe that you can do a similar trick with getting a bundle of $n$-th homotopy groups over $X$ if $X$ admits a universal cover, and that this was to be in the Dyer and Eilenberg book on algebraic topology.
[UPDATE: As Tom Goodwillie points out, this is much more complicated than necessary and misunderstands the line of argument that he had in mind. Still, it has some interesting features so I will leave it here.]
Let $\mathcal{M}$ be the category of connected smooth manifolds and smooth maps, let $\mathcal{M}_1$ be the subcategory with the same objects whose morphisms are the diffeomorphisms, and let $J\colon\mathcal{M}_1\to\mathcal{M}$ be the inclusion. Suppose we have a functor $U\colon\mathcal{M}_1\to\mathcal{M}$ and a natural map $p\colon UM\to JM$ that is a universal cover for all $M$. Consider $S^1$ as the usual subspace of $\mathbb{C}$, and choose a point $a\in p^{-1}\{1\}\subset U(S^1)$. For $z\in S^1$ we can define $\mu_z\in\mathcal{M}_1(S^1,S^1)$ by $\mu_z(u)=zu$, and then define $s(z)=U(\mu_z)(a)\in U(S^1)$. This defines a section $s$ of the map $p\colon U(S^1)\to S^1$. If we make enough additional assumptions to ensure that $s$ is continuous, then we arrive at a contradiction.
I think that in fact no additional assumptions are needed, but that needs a slightly different approach. We can identify $S^1$ with $\mathbb{R}P^1$, and then we have an action of the group $G=PSL_2(\mathbb{R})$. Let $H$ be the upper triangular subgroup, which is the stabiliser of the basepoint $1\in S^1$. For $h\in H$ there is a unique $h'\colon U(S^1)\to U(S^1)$ with $ph'=hp$ and $h'(a)=a$. The map $Fh$ need not obviously fix $a$ so it need not coincide with $h'$, but it must have $Fh=\phi(h)\circ h'$ for some deck transformation $\phi(h)$. The group of deck transformations can be identified with $\pi_1(S^1,1)=\mathbb{Z}$, and $H$ acts on this in a natural way (independent of the supposed existence of $U$). Using the connectivity of $H$ we see that this action is trivial. I think it follows that $\phi\colon H\to\mathbb{Z}$ is a homomorphism, but any element $h\in H$ has $n$'th roots for all $n>0$, and this forces $\phi$ to be trivial, so $Fh=h'$ for all $h$. This proves that $Fh$ depends continuously on $h$ for $h\in H$. Moreover, one can find $h_z,k_z\in H$ such that the entries are rational functions of $z$ and $\mu_z=h_z\mu_{-1}k_z$. It follows that $F(\mu_z)$ depends continuously on $z$ except possibly at finitely many values of $z$. These possible exceptions can then be removed by an auxiliary argument with the group structure.
If you want something functorial and base-point-independent, one option is the following $\widetilde X$ bundle over $X$. It combines all the base-point-dependent covering spaces into one gadget.
Let $C(X)$ be the space of all maps $I \to X$, modulo homotopy-rel-end-points. Let $p:C(X) \to X$ be the evaluation at the initial endpoint of $I$. It's easy to see that $p^{-1}(x)$ is the usual universal cover $\widetilde X_x$ contructed using the base point $x\in X$. So $p : C(X) \to X$ is an $\widetilde X$ bundle over $X$.
The assignment $$ X \; \mapsto \; (p : C(X) \to X) $$ is functorial in $X$.
Here is another attempt at pinning down the meaning of "canonical" in reference to Tom's answer.
Let $X$ be a nice space (connected, locally path-connected and semi-locally simply connected).
Let $\pi_X$ be the fundamental groupoid of $X$: this is a category whose objects are points $x\in X$, where a morphism $x\to y$ is a homotopy class of path fixing the endpoints.
Let $U_X$ be the groupoid of universal covers: an object is a universal cover $X_1 \to X$ and a morphism $X_1 \to X_2$ is an isomorphism of covers over the identity map of $X$.
There is a functor $$f:\pi_X\to U_X$$ (i.e., a homomorphism of groupoids) given by the usual construction of a universal cover. Then $f$ is an equivalence of categories (covering space theory).
Let $$g: U_X \to \pi_X$$ be its adjoint (which is defined up to unique isomorphism).
This means for any $\tilde X\in U_X$, with $g(\tilde X) = x\in X$ we have a preferred isomorphism $$ f(x) \cong \tilde X\, . $$
In other words, a universal cover determines a basepoint and a basepoint determines a universal cover.
