When one studies equivariant geometry, then it is sometimes the case that one can pass to a smaller space with the action of a smaller group, and everything equivariant you can compute will be the same. In the case that the quotient by the original group action exists, then this just reduces the space to the non-equivariant setting, namely equivariant for the trivial group. Groupoids are one way to encode equivariant geometry, via *action groupoids* (which are a good first-pass example to keep in mind at all times), but can permit much more general types of equivariance, even cases where it doesn't seem like it should be called that, for instance, when one has an equivalence relation on a space (the quotient might not exist, for instance, when studying leaf spaces of foliations).

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 $G$.

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.

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.

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