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Dec 5, 2018 at 21:43 vote accept Praphulla Koushik
Dec 4, 2018 at 21:05 comment added Praphulla Koushik @Qfwfq Oh.. Ok Ok.. Do you have any views to share on my question?
Dec 4, 2018 at 20:52 comment added Praphulla Koushik This is one of the data that need for a $\mathcal{G}-\mathcal{H}$ bibundle namely a $\mathcal{H}$ bundle whose base is $\mathcal{G}_0$. Now, some how this $Q$ has a $\mathcal{G}$ action that is compatible with $\mathcal{H}$ action giving a $\mathcal{G}-\mathcal{H}$ bibundle..
Dec 4, 2018 at 20:49 comment added Praphulla Koushik After reading repeatedly (It is only my problem that I did not get it immediately), I could understand (guess) something in last but one paragraph. correct me if I am wrong. Suppose I have a map $B\mathcal{G}\rightarrow B\mathcal{H}$. Consider the map of stacks $\underline{\mathcal{G}_0}\rightarrow B\mathcal{G}$. Then consider the composition $\underline{\mathcal{G}_0}\rightarrow B\mathcal{G}\rightarrow B\mathcal{H}$. This composition $\underline{\mathcal{G}_0}\rightarrow B\mathcal{H}$ should correspond to a $\mathcal{H}$ bundle over $\mathcal{G}_0$ say $Q\rightarrow \mathcal{H}_0$.
Dec 4, 2018 at 20:32 comment added Bertram Arnold What I meant was that if $B\mathcal G$ was a manifold, we could use the (2-)Yoneda lemma to compute the groupoid of natural transformations. Furthermore since $B\mathcal H$ is a stack we could equally well describe this groupoid as descent data for any cover of $B\mathcal G$. Now it is not a manifold, but it has a cover by a manifold, so the groupoid of descent data still makes sense, and it can be computed to be the groupoid of bibundles.
Dec 4, 2018 at 20:22 comment added Qfwfq @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.
Dec 4, 2018 at 19:53 comment added Praphulla Koushik @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..
Dec 4, 2018 at 19:51 comment added Praphulla Koushik @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 $B\mathcal{G}\rightarrow B\mathcal{H}$..
Dec 4, 2018 at 19:28 comment added Qfwfq " 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 $\mathcal{H}$)?
Dec 4, 2018 at 17:30 comment added Praphulla Koushik 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?
Dec 4, 2018 at 17:06 history answered Bertram Arnold CC BY-SA 4.0