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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.

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.

Part of 10.5.8 of Topology and Groupoisds 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.

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.

Part of 10.5.8 of Topology and Groupoisds 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.

Part of 10.5.8 of Topology and Groupoisds 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.