Topology of fundamental groupoid. Given fundamental groupoid $\Pi_1(S^1)$ of the circle, how can one define a topology on it? The information on nlab did little help other than the fact that it can be done since $S^1$ is path-connected. Set theoretically we can make the association with $S^1 \times \mathbb{R}$ but how can we define a topology on $\Pi_1(S^1)$ so that we have a homeomorphism between them?
 A: In order to make the groupoid $\Pi_1(X)$ into a topological groupoid we need to put a topology on the set of objects of $\Pi_1(X)$ and a topology on the set of morphisms of $\Pi_1(X)$. This is an example of a category internal to $\text{Top}$.
The set of objects already has a topology since objects are just points in $X$.
We can define a topology on the set of homotopy classes of paths in $X$ by defining an open set $S_{[\gamma],U,V}$ around $[\gamma]$ to be the set of all $[\eta * \gamma * \phi]$ where $\phi$ is completely contained in some open $U \subset X$ and $\eta$ is completely contained in some open $V \subset X$.
Assuming that the composition of paths is defined, i.e that $\eta(0) = \gamma(1)$ etc.
The topology on the set of homotopy classes of paths in $X$ is the topology generated by all the $S_{[\gamma],U,V}$.
This is similar to how the set of homotopy classes of paths beginning at a base point $x_0 \in X$ is topologized in Hatcher to construct the universal cover.
A: For CW complexes (like $S^1$), the topology on the fundamental groupoid will be discrete, i.e. it is just a usual groupoid. The idea for an arbitrary space is that the morphism sets in the fundamental groupoid is a quotient of the path space from point a to point b. Quotients have a natural topology on them, so this is a reasonable topology to put on the homotopy classes of paths from point a to point b. The issue is that composition of paths does not give a continuous map, so these topologies do not make the fundamental groupoid into a topological groupoid.
These issues can be fixed by hand, and this is done so by Jeremy Brazas (who is pretty active on these sites) Here, in his paper "The fundamental group as a topological group." Of course, this is not directly what you ask, but the techniques will work for the groupoid.
As an aside, a very reasonable way to have topological groupoids interact with CW complexes in an interesting way, is to not take things up to homotopy. The fundamental infinity groupoid of $X$ has its object space the points of $X$ and the morphism spaces the (Moore) path spaces of $X$. Then by taking path components of morphism spaces, we can recover the usual fundamental groupoid. This object is very important and studying it is part of the foundations of infinity category theory.
A: Given a connected, locally path-connected, and semi-locally simply connected space $M$ (so that $M$ has a universal covering space), the fundamental groupoid $\Pi(M)$ (viewed here as simply being the set of path homotopy classes in $M$) can be topologized as follows.
Let $\pi:\tilde{M}\to M$ be the universal cover and $\text{Aut}(\pi)$ be the deck transformation group.
Then $\text{Aut}(\pi)$ acts diagonally on $\tilde{M}\times \tilde{M}$, and the quotient space $(\tilde{M}\times \tilde{M})/\text{Aut}(\pi)$ is in one-to-one correspondence with $\Pi(M)$, so we may give $\Pi(M)$ this quotient topology.
If additionally $M$ is a smooth manifold and $\tilde{M}$ has the unique smooth structure making $\pi$ smooth, then the action of $\text{Aut}(\pi)$ on $\tilde{M}$ is smooth, free, and proper.
The same is true of the diagonal action of $\text{Aut}(\pi)$ on $\tilde{M}\times\tilde{M}$, so the quotient manifold theorem implies that $\Pi(M)= (\tilde{M}\times \tilde{M})/\text{Aut}(\pi)$ is a smooth manifold and $\tilde{M}\times\tilde{M}\to \Pi(M)$ is a smooth covering map.
A: Grupoid of $Y$ is the factor of the set of all map $[0,1]\to Y$ modulo homotpies. On the set of all maps between two toplogical spaces $Map(X,Y)$ there is so-called compact-open toplogy:https://en.wikipedia.org/wiki/Compact-open_topology
Then take quotient toplogy.
Definition of compact-open topology:
Let $K\subset X$ be a compact set and $U\subset Y$ be some open. Let $V(K,U)$ be the set of all maps $X\to Y$ such that $f(K)\subset U$. Then take the smallest topology such that all $V(K,U)$ are open.
