classifying space induces a equivalence of categories between PBun$_G(M)$ and $\Pi(M,BG)$ for finite groups $G$ Let $G$ be a finite group, $BG$ its classifying space and M a manifold. Then it is mentioned in https://arxiv.org/abs/1705.05171 (Remark 2.3 d) that there is an equivalence of categories
$$
\Pi (M,BG) \cong PBun_G(M)
$$
between the groupoid of maps from $M$ to $BG$ (with homotopy-classes of homotopies as morphisms) and the groupoid of principal $G$-bundles over $M$. How can one prove that? Where in the litterature can I find a proof of that?
Yet I could only find proofs that homotopy classes of maps correspond to isomorphism classes of principal bundles. But in order to prove that it gives an equivalence of categories I also need that the morphisms correspond to each other.
That they correspond to each other is also mentioned in https://www.uni-due.de/~hm0002/stacks.pdf (In the beginning of chapter 1), but no proof is given.
Thank you in advance for your help
 A: To have an equivalence of groupoids we need an equivalence on components and on automorphism groups of each component. The equivalence on components is the classical story. The reference I like best for this is Stephen Mitchell's notes https://www3.nd.edu/~mbehren1/18.906/prin.pdf
To get equivalence on automorphism groups is not difficult. Suppose $H: X \times I \rightarrow BG$ is a homotopy from the map $f: X \rightarrow BG$ to itself. Then this can instead be written as $H: X \times S^1 \rightarrow BG$ with the restriction to $X \times \{*\}$ equal to $f$.
We may pull back the universal bundle to get a vector bundle over $X \times S^1$. Because our group is discrete, by going in the counter-clockwise direction we have a unique path associated to $x \in X$ starting at $(x,\{*\})$ and returning to $(x',\{*\})$ for some $x' \in X$. Think about what happens on the boundary of the Mobius strip.
This assignment $x \rightarrow x'$ then gives an automorphism of the pullback bundle along $f$. Hence, we have a map from the self homotopies of a map to the automorphism group of the pullback bundle. To see that this is an isomorphism, we construct the inverse map.
Given an automorphism of the principal bundle $P \rightarrow X$, pullback along the projection to get a principal bundle over $X \times I$. Now we may glue along the automorphism to get a principal bundle over $X \times S^1$. This is classified by a map $X \times S^1 \rightarrow BG$ which gives rise to a self homotopy of the map classifying $P \rightarrow X$. It is not hard to check these are inverse.
Notably, the first map requires discreteness and the second does not.
