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

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 $$\mathcal{H}$$.

• What do you mean by an element of a stack? For me a stack is a sheaf of groupoids, and the principal bundle $t:\mathcal G_1\to\mathcal G_0$ is an object in the groupoid $B\mathcal G(\mathcal G_0)$. A morphism of stacks $B\mathcal G\to\mathcal BH$ sends this to an object of $B\mathcal H(G_0)$, and if you present the morphism as a functor $F_i:\mathcal G_i\to\mathcal H_i$, you can check that this object is the principal $\mathcal H$-bundle $\mathcal H_1\times_{t,\mathcal H_0,F_0}\mathcal G_0$ (the map $F_i$ and functoriality is used to turn this into a principal bundle). – Bertram Arnold Dec 4 '18 at 11:53
• I think you mean to say "[---] If you present the morphism as a functor $F:\mathcal{G}\rightarrow \mathcal{H}$ [---]"... @BertramArnold – Praphulla Koushik Dec 4 '18 at 12:42
• There is something here which does not make sense. As any map of stacks $B\mathcal{G}\rightarrow B\mathcal{H}$ is fiber preserving, it takes an object $t:\mathcal{G}_1\rightarrow \mathcal{G}_0$ in $B\mathcal{G}$ to some thing of the form $Q\rightarrow \mathcal{G}_0$ which can not be $t:\mathcal{H}_1\rightarrow \mathcal{H}_0$. I was having in mind the map of stacks coming from a Lie groupid extension (which is simply a morphism of Lie groupoids with same base i.e., $\mathcal{G}_0=\mathcal{H}_0$ with some extra condition on $\phi:\mathcal{G}_1\rightarrow \mathcal{H}_1$). – Praphulla Koushik Dec 4 '18 at 13:13
• In that case, object $t:\mathcal{G}_1\rightarrow \mathcal{G}_0$ goes to some thing of the form $Q\rightarrow \mathcal{G}_0$ in $B\mathcal{H}$. As $\mathcal{H}_0=\mathcal{G}_0$, object $t:\mathcal{G}_1\rightarrow \mathcal{G}_0$ goes to some thing of the form $Q\rightarrow \mathcal{H}_0$ in $B\mathcal{H}$. Now, I can ask if this $Q\rightarrow \mathcal{H}_0$ is same as that of the unit element $\mathcal{H}_1\rightarrow \mathcal{H}_0$ in $B\mathcal{H}$. This is what I meant to ask. I am editing the question so that it makes some sense. @BertramArnold – Praphulla Koushik Dec 4 '18 at 13:13
• As you said a general morphism of stacks can't preserve "unit" principal bundles. If both stacks are presented by groupoids and the morphism is presented by a functor it sends the unit bundle of $\mathcal G$ to the pullback of the unit bundle of $\mathcal H$ along the object map of the functor (which should be something like the formula at the end of my previous comment), so in case of a Lie groupoid extension it does preserve unit bundles. All of this should more or less follow from the definitions, in particular how a Lie functor defines a fibered functor between the presented stacks. – Bertram Arnold Dec 4 '18 at 13:48

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.

• When you have a map of stacks $\mathcal{D}\rightarrow \mathcal{C}$ you want to think $\mathcal{C}$ as stack associated to a manifold $M$, some denote this by $\underline{M}=B\{M\rightrightarrows M\}$ and you want to think $\mathcal{D}$ as stack associated to a Lie groupoid $\mathcal{G}$ which is denoted by $B\mathcal{G}$... Is this what you mean in last paragraph? – Praphulla Koushik Dec 4 '18 at 17:30
• " 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 $B\mathcal G$" - Wait.. this would be a 2-version of Yoneda I guess? And by the way what is $B\mathcal{H}(B\mathcal{G})$ (if not tautologically the Hom groupoid $\mathrm{Stacks}(B\mathcal{G},B\mathcal{H})$ of $1$-morhpisms (="Lie functors") from $B\mathcal{G}$ to $\mathcal{H}$)? – Qfwfq Dec 4 '18 at 19:28
• @Qfwfq Even I do not understand what is "evaluation of $B\mathcal{H}$ on $B\mathcal{G}$"... The $2$-version of Yoneda lemma I am aware of is (which you can find for example in page $23$ of Orbifolds as Stacks by Eugene Lerman) that, "given a manifold $M$ and a Lie groupoid $\mathcal{G}$, the category of maps/natural transformations from $\underline{M}$ to $B\mathcal{G}$ is evaluation of $B\mathcal{G}$ on $M$" i.e., the fiber $B\mathcal{G}(M)$ which are just principal $\mathcal{G}$ bundles over $M$.. I do not know if this can be generalized to the case $B\mathcal{G}\rightarrow B\mathcal{H}$.. – Praphulla Koushik Dec 4 '18 at 19:51
• @Qfwfq appropriate thing on other side would be "Evaluation of $B\mathcal{H}$ on $B\mathcal{G}$" just like in case of $\underline{M}\rightarrow B\mathcal{G}$ it was evaluation of $B\mathcal{G}$ on $M$... What I know is there is a correspondence between maps of the form $B\mathcal{G}\rightarrow B\mathcal{H}$ and what are known as $\mathcal{G}-\mathcal{H}$ bibundles.. – Praphulla Koushik Dec 4 '18 at 19:53
• @PK: yep, that one you quote may indeed be called a "2-Yoneda". I was talking about a (hypothetical) version with objects of 2-categories (instead of objects of 1-categories) everywhere. – Qfwfq Dec 4 '18 at 20:22