unit element under map of stacks $B\mathcal{G}\rightarrow B\mathcal{H}$

Question

Let $\mathcal{G}$ be a Lie groupoid. The target map $t:\mathcal{G}_1\rightarrow \mathcal{G}_0$ is a principal $\mathcal{G}$ bundle.

This article Orbifolds as Stacks? by Eugene Lerman calls (in page $11$) this particular principal bundle to be the unit principal $\mathcal{G}$ bundle. So, for a Lie groupoid $\mathcal{G}$, this $\mathcal{G}$ bundle is a special element in $B\mathcal{G}$.

Let $\mathcal{G}$ and $\mathcal{H}$ be Lie groupoids. For $B\mathcal{G}$, I have a special element $t:\mathcal{G}_1\rightarrow \mathcal{G}_0$. For $B\mathcal{H}$, I have a special element $t:\mathcal{H}_1\rightarrow \mathcal{H}_0$. We can ask the follwing question.

Are "the unit elemnts" preserved by an arbitrary map of stacks $F:B\mathcal{G}\rightarrow B\mathcal{H}$ i.e., do we always have $F(t:\mathcal{G}_1\rightarrow \mathcal{G}_0)=t:\mathcal{H}_1\rightarrow \mathcal{H}_0$ for any map of stacks $F:B\mathcal{G}\rightarrow B\mathcal{H}$.

This question does not make sense. As $F$ preserves fibers, $F(t:\mathcal{G}_1\rightarrow \mathcal{G}_0)$ would be a principal $\mathcal{H}$ bunle of the form $Q\rightarrow \mathcal{G}_0$. So, $F(t:\mathcal{G}_1\rightarrow \mathcal{G}_0)$ which is of the form $Q\rightarrow \mathcal{G}_0$ can not be the principal $\mathcal{H}$ bundle $t:\mathcal{H}_1\rightarrow \mathcal{H}_0$.

We have principal $\mathcal{H}$ bundles $Q\rightarrow \mathcal{G}_0$ and $\mathcal{H}_1\rightarrow \mathcal{H}_0$. As base spaces are different there is no way (in general) to compare these two principal $\mathcal{H}$ bundles.

Suppose we are in a situation where $B\mathcal{G}\rightarrow B\mathcal{H}$ is coming from a morphism of Lie groupoids $\mathcal{G}\rightarrow \mathcal{H}$. Realising that there is an obvious map between base spaces of these two bundles namely $\phi:\mathcal{G}_0\rightarrow \mathcal{H}_0$ makes the question of looking for some relation between $Q\rightarrow \mathcal{G}_0$ and $\mathcal{H}_1\rightarrow \mathcal{H}_0$ more specific. The question would then be,

Is $Q\rightarrow \mathcal{G}_0$ same the pull back of $t:\mathcal{H}_1\rightarrow \mathcal{H}_0$ along $\phi:\mathcal{G}_0\rightarrow \mathcal{H}_0$.

In a very special case when this morphism of Lie groupoids $\phi:\mathcal{G}\rightarrow \mathcal{H}$ is a Lie groupoid extension i.e., when $\phi:\mathcal{G}_0\rightarrow \mathcal{H}_0$ is an identity map (and some thing extra), we can ask

Is $Q\rightarrow \mathcal{G}_0$ same thing as $t:\mathcal{H}_1\rightarrow \mathcal{H}_0=\mathcal{G}_0$.

I could not see why this is true from definition of map of stacks but I feel this should be the case. Any comments are welcome.

Edit : A stack (over the category of manifolds $\text{Man}$) for me is a category $\mathcal{D}$ along with a functor $\mathcal{D}\rightarrow \text{Man}$ such that it is a category fibered in groupoids and some extra conditions. By an element of stack I mean an element (object) in the category $\mathcal{D}$.

Consider the special case when this map of stacks $B\mathcal{G}\rightarrow B\mathcal{H}$ is coming from a morphism of Lie groupoids $\mathcal{G}\rightarrow \mathcal{H}$. I will recall how one gets $B\mathcal{G}\rightarrow B\mathcal{H}$ from $\mathcal{G}\rightarrow \mathcal{H}$.

We first construct a $\mathcal{G}-\mathcal{H}$ bibundle given $\phi:\mathcal{G}\rightarrow \mathcal{H}$. We consider principal $\mathcal{H}$ bundle $t:\mathcal{H}_1\rightarrow \mathcal{H}_0$, pull it back along $\phi:\mathcal{G}_0\rightarrow \mathcal{H}_0$ to get a principal $\mathcal{H}$ bundle, now with the base $\mathcal{G}_0$ which is precisely $\mathcal{G}_0\times_{\mathcal{H}_0}\mathcal{H}_1\rightarrow \mathcal{H}_0$. The manifold $\mathcal{G}_0\times_{\mathcal{H}_0}\mathcal{H}_1$ has an action of $\mathcal{G}$ on it, given by $g.(x,h)=(t(g),\phi(g)h)$. With this action, it becomes a $\mathcal{G}-\mathcal{H}$ bibundle.

This $\mathcal{G}-\mathcal{H}$ bibundle gives morphism of stacks $B\mathcal{G}\rightarrow B\mathcal{H}$ the map is given by composition of bibundle. The object $\mathcal{G}_1\rightarrow \mathcal{G}_0$ is mapped to $\left(\mathcal{G}_1\times_{\mathcal{G}_0}(\mathcal{G}_0\times_{\mathcal{H}_0}\mathcal{H}_1)\right)/\mathcal{G}_1\rightarrow \mathcal{G}_0$ which comes from following diagram

We have $\left(\mathcal{G}_1\times_{\mathcal{G}_0}(\mathcal{G}_0\times_{\mathcal{H}_0}\mathcal{H}_1)\right)/\mathcal{G}_1= \mathcal{G}_0\times_{\mathcal{H}_0}\mathcal{H}_1$.

So, the principal bundle is $\mathcal{G}_0\times_{\mathcal{H}_0}\mathcal{H}_1\rightarrow \mathcal{G}_0$.

If $\mathcal{G}\rightarrow \mathcal{H}$ is a Lie groupoid extension, then we have $\mathcal{H}_0=\mathcal{G}_0$ and $\mathcal{G}_0\times_{\mathcal{H}_0}\mathcal{H}_1=\mathcal{H}_1$. So, $\mathcal{G}_0\times_{\mathcal{H}_0}\mathcal{H}_1\rightarrow \mathcal{G}_0$ is just the target map $t:\mathcal{H}_1\rightarrow \mathcal{H}_0$.

So, if $F:B\mathcal{G}\rightarrow B\mathcal{H}$ is given by a morphism of Lie groupoids $\mathcal{G}\rightarrow \mathcal{H}$ which is a Lie groupoid extension, then $F(t:\mathcal{G}_1\rightarrow M)=t:\mathcal{H}_1\rightarrow M$.

Otherwise, it does not make sense to ask if $F$ takes unit principal bundle of $\mathcal{G}$ to unit principal bundle of $\mathcal{H}$.

Answer 1

This is more of a long comment, but since the question has been answered in the last edits I will post it as an answer.

There are multiple perspectives on the stack presented by a Lie groupoid $\mathcal G$: One is the fibered category of principal bundles which you use, another one is the sheafification of the presheaf of groupoids which sends $X$ to the groupoid with objects maps from $X$ to $\mathcal G_0$ and morphisms maps from $X$ to $\mathcal G_1$. These correspond to trivial principal bundles; in the stackification we have to replace our test manifold by a cover $U$ and consider trivial principal bundles on $U$ together with descent data on $U\times_X U$. You can check that these glue to a principal bundle on $X$. (If $\mathcal G = pt//G$, this is just the description of principal $G$-bundles by cocycles.)

Both descriptions are equivalent, and the principal bundle description is arguably nicer - it involves a short list of data and relations, whereas the stackification of the prestack requires you to allow all possible covers, then identify descent data on different covers when there is a common refinement, ... However this description only works well if we map into the stack $B\mathcal G$ (for instance, when evaluating it on a test manifold). When we map out of it, we usually use the cover coming from the Lie groupoid presentation:

If $\mathcal G$ and $\mathcal H$ are two Lie groupoids, one can ask what the groupoid of natural transformations between $B\mathcal G$ and $B\mathcal H$ is. By the Yoneda lemma, this should correspond to the evaluation of $B\mathcal H$ on $B\mathcal G$, and since the unit principal bundle defines a morphism $\mathcal G_0\to B\mathcal G$ which is a cover, this evaluation is just given by objects in $B\mathcal H(\mathcal G_0)$ together with an isomorphism between the two ways of pulling this object back to $\mathcal G_1$. You can check that this recovers the notion of a bibundle between Lie groupoids - the object in $B\mathcal H(\mathcal G_0)$ defines the right $\mathcal H$-bundle structure, and the isomorphism defines the left $\mathcal G$-bundle structure. From this description you can immediately derive all formulas for bibundles, e.g. the composition of bibundles, the bibundle associated to a functor, ...

Lastly, I think that while it's important to have a rigourous mathematical framework for stacks, for which fibered categories are certainly a good candidate, it's also important to have intuition about them, and for this I usually pretend that my stack is $BG$ when I map into it and the Cech groupoid of a cover of some manifold when I map out of it.