# Necessity/Motivation for generalised homomorpisms

I am reading Ieke Moerdijk's article "Orbifolds as Groupoids : an Introduction".

In that notes author defines a notion of generalized map between Lie groupoids.

Let $\mathcal{G}$ and $\mathcal{H}$ be Lie groupoids. By a generalized map from $\mathcal{G}$ to $\mathcal{H}$ we mean a homomorphism of Lie groupoids $\mathcal{G}'\rightarrow \mathcal{H}$ where $\mathcal{G}'$ is a morita equivalent groupoid to $\mathcal{G}$.

I do not understand the necessity/motivation behind introducing this notion of generalized map.

What can one expect to do by considering maps of this generality?

Even the notion of Morita equivalent is little extra that what one would expect.

Two Lie groupoids $\mathcal{G}$ and $\mathcal{G}'$ are said to be Morita equivalent if there exists a third groupoid $\mathcal{H}$ and equivalences (a morphsim of Lie groupoids said to be an equivalence if somthing happens) $$\mathcal{G}\xleftarrow{\phi} \mathcal{H}\xrightarrow{\phi'} \mathcal{G}'.$$

I am not able to see the necessity of considering this general notion of equiavelnce.

Edit : This is just to bump the question so that it gets some attention and an answer/comment.

Edit: Orbifold definition that I am familiar with is the following.

Definition : Let $X$ be a locally compact Hausdorff space. An orbifold structure on $X$ is given by an orbifold groupoid $\mathcal{G}$ and a homeomorphism $f:|\mathcal{G}|\rightarrow X$. If $\mathcal{H}\rightarrow\mathcal{G}$ is a equivalnece, then $|\phi|:|\mathcal{H}|\rightarrow |\mathcal{G}|$ is a homeomorphism and the composition $f\circ|\phi|:|\mathcal{H}|\rightarrow |\mathcal{G}|\rightarrow X$ is viewed as defining an equivalent orbifold structure on $X$. An orbifold $\underline{X}$ is a space $X$ equipped with an equivalnece class of orbifold structures.

• This is clearer from the point of view of stacks: morally two groupoids are Morita equivalent iff they present the same stack. Jun 3, 2018 at 17:36
• @DenisNardin I only know little about stacks(a category fibered in groupoid with some extra conditions) and little about groupoids.. I dont know how a groupoid present a stack.. can you say little more. Jun 3, 2018 at 17:46
• I suggest also looking at Dorette Pronk's work in this area, eg Orbifolds, sheaves and groupoids (pdf), or arxiv.org/abs/1606.04439, or this by Tommasini: doi.org/10.1016/j.topol.2011.11.043 Jun 4, 2018 at 2:49
• @DavidRoberts Thank you :) I will see that.. Jun 4, 2018 at 4:10
• @DavidRoberts I still could not get her thesis. Can you write down little more about why we need generalised morphism s here. Thanks. :) Jun 23, 2018 at 3:42

Let me attempt a very simple-minded answer.

Say your objects of interest are orbifolds. You have an orbifold V and you want to describe it through a groupoid, usually an action groupoid $G\ltimes M \rightrightarrows M$. This means that you have an action of $G$ on $M$ and that $V\simeq G\backslash M$, i.e. $V$ is the orbit space of the action groupoid.

Then you may think that a map between two such orbifolds $V_1$ and $V_2$ is a groupoid morphism between the corresponding action groupoids.

But this does not work because there are many ways in which you can construct an action groupoid with orbit space $V$. There are many different ways in which you can see an orbifold as quotient of a manifold. And you cannnot expect to lift any orbifold map to any possible such presentation.

Morita equivalence is the equivalence relation that guarantees you that two different groupoids have the same orbit space. Thus that they, in a way, correspond to different ways of presenting the same orbifold.

Generalized morphism are the kind of correspondence that you can establish between one presentation of the first orbifold and one presentation of the other orbifold.

This is the reason why you need the generality of Morita equivalence and generalized morphisms in the context of orbifolds.

Nardin comment is a very terse way to say this: an orbifold (more generally a stack) is described by a Morita equivalence class of groupoids.

From this point of view it is very relevant to understand which groupoid invariants are, in fact, Morita invariant. Those will define orbifold (stack) invariants.

• I do not completely understand... For an orbifold, there may be two presentations.. This is ok but I don’t understand “And you cannnot expect to lift any orbifold map to any possible such presentation.” Jun 25, 2018 at 10:23
• Whatever "an orbifold map" is (and if you know what it is you know that it is not completely easy to define it) if $V_i$ is the orbit space of $G_i\ltimes M_i\rightrightarrows M_i$ there is no reason to expect that a map $V_1\to V_2$ lifts to a groupoid map $G_1\ltimes M_1$ to $G_2\ltimes M_2$.Maps between orbit spaces of groupoids does not imply the existence of a groupoid morphism. Jun 25, 2018 at 10:33
• I think I understand to some extent. Please give some time, I will respond. Jun 25, 2018 at 13:44
The idea that one can pass to a different presentation of what is essentially the same geometry means that one wants some notion of isomorphism (or more generally, equivalence) between two such presentations. But it is simply not true that in the category (or even the 2-category) of Lie groupoids and functors (and natural transformations) that there is an equivalence between two such presentations. For instance, consider the following two examples, starting from a manifold $X$ with an action by a Lie group $G$.
1. Assume $G$ has in fact a compact normal subgroup $K$ with $H:=G/K$ and such that $K$ acts freely on $X$. Then $M:=X/G$ has a canonical $H$-action. There is a (surjective) functor $X//G \twoheadrightarrow M//H$ between the respective action groupoids. Unless there is a section of $X\to M$, this functor is not an equivalence.
2. Assume that there is a global slice $Y\subset X$ (so for every point $x\in X$ there is $g\in G$ with $gx\in Y$), with the strong property that there is a subgroup $L\lt G$ and if $g\in G$ is such that $gy\in Y$ (for $y\in Y$), then $g\in L$. Then there is an inclusion of action groupoids $Y//L \hookrightarrow X//G$. This is not an equivalence unless there is an equivariant retraction $X\to Y$.
In both of these cases, the equivariant geometry is identical, in the sense that they are both presentations of a possibly non-existent quotient $X/G$ (meaning: non-existent in the category of manifolds). Even worse, one might have both of these examples happening at the same time, so that one has two action groupoids presenting the same equivariant geometry, but no functor in either direction. For instance, one could have $$X//G \twoheadrightarrow M//H \hookleftarrow Y//L$$ Generalised morphisms of all kinds are meant to be the "fix" for this situation. Now it just so happens that Lie groupoids (and other types of 'geometric' groupoids) also give rise to objects called stacks (in a functorial way), and there is the pleasant fact that the functors of Lie groupoids that should be equivalences, but aren't, are sent to equivalences of stacks. It's not important in a first treatment to know what a stack is, but essentially orbifolds, via groupoids, are stacks, and maps between orbifolds are maps between the corresponding stacks. And, maps between stacks arising from groupoids correspond precisely to generalised morphisms of groupoids, be they anafunctors or bibundles (or Hilsum-Skandalis maps, or ....).
One way to define generalised morphisms from $A$ to $C$ is as spans $A \leftarrow B \to C$ of functors, but where the functor $B\to A$ comes from a special class of functors that should count as equivalences, but aren't. Not any such functor should be used (for Reasons :-) but those whose object component is a surjective submersion. Such spans are called anafunctors, and were originally introduced by Makkai, and further developed for more general settings by Toby Bartels. Right principal bibundles are the other main approach, and these correspond closely to a restricted class of anafunctors, called saturated. They are not themselves spans of groupoids, but one can reconstruct such a span from them.