Morphism of Lie groups $\theta:G\rightarrow H$ giving an equivalence of categories $BG\rightarrow BH$?

Question

Given a morphism of Lie groups $ \theta:G\rightarrow H$  and a principal $G$ bundle $ \pi:P\rightarrow M$ there are (at least) two ways to assign a principal $ H$ bundle.

1. See that the morphism of Lie groups $ \theta:G\rightarrow H$ gives an action of $ G$ on $ H$ by $ g.h=\theta(g).h$. Given an action of $ G$ on manifold (Lie group in this case) $ H$ there is an associated fibre bundle $ P\times_G H\rightarrow M$ with fibre $ H$. This gives a principal $ H$ bundle.
2. For principal bundle $ \pi:P\rightarrow M$, we can find an open cover $ \{U_\alpha\}$ of $ M$ and  (transition) maps $ g_\alpha g_\beta:U_{\alpha\beta}\rightarrow G$ satifsying the cocycle condition $ g_{\alpha\beta}g_{\beta\gamma}=g_{\alpha\gamma}$ on $ U_\alpha\cap U_\beta\cap U_\gamma$. Then the compositions $ \tau_{\alpha\beta}=\theta\circ g_{\alpha\beta}:U_{\alpha\beta}\rightarrow G\rightarrow H$ also satifies the cocycle condition $ \tau_{\alpha\beta}\tau_{\beta\gamma}=\tau_{\alpha\gamma}$ on $ U_\alpha\cap U_\beta\cap U_\gamma$. One can then produce a principal $ H$ bundle over $ M$ given this open cover $ \{U_\alpha\}$ of $ M$ and smooth maps $ \tau_{\alpha\beta}:U_\alpha\cap U_\beta\rightarrow H$ satisfying the cocycle condition. This gives a principal $ H$ bundle.

It is a good exercise (that I have not tried) to check that principal $ H$ bundles obtained from above two methods are (naturally) isomorphic.

Given a Lie group $ G$, let $ BG$ denote the category of principal $ G$ bundles. Objects are principal $ G$ bundles and morphisms are $ G$-equivariant morphisms.

Given a morphism of Lie groups $ \theta:G\rightarrow H$, above construction gives a functor (at the level of objects) $ B\theta:BG\rightarrow BH$. It is not difficult to see that, a $ G$-equivarint map induce a $ H$-equivariant map. This gives a functor.

I am trying to understand what can we say about $\theta:G\rightarrow H$ if we know that $B\theta:BG\rightarrow BH$ is an equivalence of categories? Does it have to be a diffeomorphism? Any comments are welcome.

Answer 1

It is a diffeomorphism, since there is an equivalence of bicategories $DifferentiableStacks \simeq LieGroupoids[W^{-1}]$ where the RHS is the bicategorical localisation of the usual 2-category of Lie groupoids a la Pronk. Because of the special nature of the domain of your $\theta$, namely it is a Lie groupoid with one object (call it $\mathbf{B}G$; note the boldface B!), then in fact $$ LieGroupoids[W^{-1}](\mathbf{B}G,\mathbf{B}H) \simeq LieGroupoids(\mathbf{B}G,\mathbf{B}H) $$ is an equivalence of categories. The latter category is isomorphic to the category whose objects are homomorphisms $G\to H$ and whose 2-arrows are elements of $H$, acting by conjugation on homomorphisms. Tracing what happens to the quasi-inverse $\phi\colon BH\to BG$ to $B\theta$ through these equivalences of categories one gets a homomorphism $\psi\colon H\to G$ such that $\psi\circ \theta$ is conjugate to the identity map on $G$, and $\theta\circ \psi$ is conjugate to the identity map on $H$. This is enough to know that $\theta$ is a diffeomorphism, since we can pre- and post-compose $\psi$ with the inner automorphisms to get an inverse for $\theta$.