To edify my understanding of fiber bundles with structure groups, I was currently trying to reconcile two classifications (in a particular case). For simplicity, I'm taking the base to be $S^1$ and the group $G$ to be discrete. Initially, I was getting a different answer from each. **Update:** The mistake there has been pointed out to me, but I'm still trying to resolve some cofusion.

We can classify principal $G$-bundles as homotopy classes of maps $S^1\rightarrow BG$. In other words, we want to take free homotopy classes of loops in $BG$, which is the same thing as conjugacy classes in $\pi_1(BG,\text{pt})$. Using $\Omega BG\simeq G$, we get $\pi_1(BG,\text{pt})\cong G$ and thus principal $G$-bundles over $S^1$ correspond to conjugacy classes in $G$.

(Initially, I forgot to take the conjugacy action when looking at homotopy classes of maps $S^1\rightarrow BG$, so I falsely assumed that principal bundles corresponded bijectively with elements of $G$. This had the same effect as working in a category of pointed spaces, similarly to how the classification of connected covering spaces involves subgroups of $\pi_1$ or their conjugacy classes, depending on whether or not our spaces are pointed.)

I see two ways of being more explicit about this correspondence, i.e. not going through the classifying space. The first is just a helpful thought process. The second is where my remaining confusion enters.

- We can look at transition functions. If we consider a fiber bundle over $S^1$ with fiber a $G$-torsor, it will be trivializable over the upper and lower semicircles and therefore given by a transition map on their intersection, which is just two elements $g_L,g_R\in G$. Two different pairs $(g_L,g_R)$ and $(h_L,h_R)$ define the same bundle if and only if they differ by some change of trivialization on the upper and lower semicircles, i.e. there exist $r_U,r_D\in G$ such that $$h_L=r_U^{-1}g_Lr_D\quad\text{and}\quad h_R=r_U^{-1}g_Rr_D.$$ This allows us to assume that every bundle is defined by a pair $(e,g)$, where $e\in G$ is the identity. But now two bundles defined by $(e,g)$ and $(e,h)$ will be isomorphic if and only if $h=r^{-1}gr$ for some $r\in G$ (we must have $r_U=r_D$ to keep $e$ in the first coordinate). This implies that the isomorphism classes of bundles correspond to conjugacy classes in $G$.
- In both formulations, the bundle corresponding to some $g\in G$ is just the mapping torus of left-multiplication by $G$. Explicitly, we take $G\times [0,1]$ and identify $(h,1)$ with $(gh,0)$ for any $h\in G$. Then the map $(x,t)\mapsto e^{2\pi it}$ clearly descends to the quotient and gives a covering of the circle. The $G$-action on the principal bundle is now just multiplication in the $G$-coordinate.

What still confuses me is the following: it seems like we can recover an explicit element $g\in G$ from the principal $G$-bundle, by taking the monodromy on a fiber. Explicitly, let $\pi:E\rightarrow S^1$ be a principal $G$-bundle, pick some $x\in \pi^{-1}(1)$ and let $\gamma:[0,1]\rightarrow E$ be the unique path with $\gamma(0)=x$ and $\pi\circ\gamma(t)=e^{2\pi i t}$. Then we get $\gamma(1)=gx$ for some $g\in G$ and this bundle is the mapping torus of left-multiplication by $G$. Initially, I thought that perhaps picking a different $x$ would lead to a conjugation of $g$, but in trying it out, I am getting the same $g$ before and after. (**Note: this was wrong.**)

In particular, I tried constructing the bundle corresponding to $(12)\in S_3$. As a covering space, this is just three copies of the connected double cover of $S^1$. But once we include the $G$-action, I get a bit confused. For now acting by $(12)$ corresponds to monodromy and therefore preserves each connected component of the total space. But acting by $(23)$ interchanges connected components! This makes it seem like the bundle *cannot* be isomorphic to one where the action of $(23)$ corresponds to monodromy. Did I compute this example wrong, or what is going on here? (**Note: this was wrong.**)