Let $G$ be a (discrete) group, and $1/G$ the corresponding groupoid with one object. Consider the diagram in (the 2-category) Groupoids with one vertex, labeled $1/G$, the one arrow from that vertex to itself, given by the identity map.  

$$ \begin{matrix} 1/G \\ {\huge \circlearrowleft} \\ \scriptstyle \mathrm{id} \end{matrix}$$
(This diagram is equivalent to the pair of parallel arrows $1/G \overset{\rm id}{\underset{\rm id}\rightrightarrows} 1/G$. Note that I am *not* filling in the loop with a 2-cell.)


A cute fact is that the ("2-") limit of this diagram in Groupoids is the action groupoid $G/G$ of the adjoint action of $G$ on itself. (See e.g. [2 limit in nLab](http://ncatlab.org/nlab/show/2-limit) or [HTT Chapter 4](http://arxiv.org/abs/math/0608040v4) for a definition of limits.)


Now, in homotopological terms, the groupoid $1/G$ looks like the classifying space ${\rm B}G$, and the above diagram looks like ${\rm B}G \times S^1$. I have the possibly-mistaken impression that limits are supposed to look like topological cones (but maybe this is because we use words like "cone" when talking about limits).


**Question:** In terms of homotopy, how should I visualize the limit cone
$$ \lim\left( \begin{matrix} 1/G \\ {\huge \circlearrowleft} \\ \scriptstyle \mathrm{id} \end{matrix} \right) \quad \begin{matrix} {\huge \to} \\ {\large \circlearrowleft \!\!\!\!\!\! \circlearrowleft} \end{matrix} \quad \begin{matrix} 1/G \\ {\huge \circlearrowleft} \\ \scriptstyle \mathrm{id} \end{matrix} $$
?


(Edits: per Quid's request in the comments, I replaced some broken images with diagrams, trying to reconstruct them from memory. $\circlearrowleft \!\!\!\!\! \circlearrowleft$ is my attempt at a doubled circle arrow, i.e. a 2-cell filling in the cone walls.)



How to interpret topologically that the equalizer in Groupoids of ${\rm id}, {\rm id}: BG \rightrightarrows BG$ is $G/G$ (adjoint action)?
You should think of the limit of that diagram as "loops in $BG$." A loop in $BG$ is a principal $G$-bundle on $S^1$. Every principal $G$-bundle on the circle comes from taking the trivial principal $G$-bundle on $[0,1]$ and identifying the fibers over $0$ and $1$ (making this the basepoint). If you like left principal bundles, this gluing map has to be right multiplication by an element $g$ of $G$. Of course, you can still do gauge transformations on the circle, and these will have the effect of conjugating $g$ by the value of the gauge transformation at the basepoint.


Thus, principal bundles on $S^1$ can be thought of as $G/G$.


