I am trying to understand what exactly is the Morita equivalence of Lie groupoids.

I am reading Ieke Moerdijk’s notes Orbifolds as groupoids.

A homomorphism $\phi:\mathcal{H}\rightarrow \mathcal{G}$ between Lie groupoids is called an equivalence (Morita equivalence) if

the composition $G_1\times_{G_0} H_0\xrightarrow{\pi_2}G_1\xrightarrow{t}G_0$ is a surjective submersion.

is a fibered product of manifolds.

Here $\mathcal{H}_1$ denotes morphism set and $\mathcal{H}_0$ denotes objects set. Similarly $\mathcal{G}_0,\mathcal{G}_1$ are denoted.

I do not really understand what exactly this says. I am trying to understand what this means in case of simple examples but did not succeed.

Notion of equivalence of categories I know is the following :

A functor $\mathcal{F}:\mathcal{C}\rightarrow\mathcal{D}$ is called an equivalnece of categories if there is another functor $\mathcal{G}:\mathcal{D}\rightarrow \mathcal{C}$ such that $\mathcal{F}\circ\mathcal{G}$ is naturally equivalent to the identity functor on $\mathcal{D}$ and $\mathcal{G}\circ\mathcal{F}$ is naturally equivalent to identity functor on $\mathcal{C}$.

Do we have something similar to this when we say Morita equivalence of Lie groupoids?

Any comments are welcome.

**Edit** : As a suggested by **Benjamin Steinberg**, I tried to see that first condition is saying $\phi$ is an essentially surjective functor and second condition is saying that $\phi$ is fully faithful.

I was able to see that first condition implies that the functor is essentially surjective. I have deleted proof here and added in comments so that question do not look big.

Now, I need to see that the second condition implies $\phi$ is full and faithful i.e., Given $x,y\in \mathcal{H}_0$ I have to see that the induced map $\mathcal{H}(x,y)\rightarrow \mathcal{G}(\phi(x),\phi(y))$ is a bijective map. I was able to see that this is surjective (proof in comments) but could not see how this is injective.

Let $\gamma,\gamma':x\rightarrow y$ be such that $\phi(\gamma)=\phi(\gamma')$. How does one prove $\gamma=\gamma'$. I can not really say what exactly does it mean to say two arrows are equal.

Any suggestion is welcome.

Once I prove that this means $\phi$ is essentially surjective and fully faithful, it gives some justification for declaring this to be a good notion of equivalence of Lie groupoids from following result.

A functor $F:A\rightarrow B$ is fully faithful and essentially surjective if an only if there is a functor $G:B\rightarrow A$ with two natural isomorphisms(natural transformations) $\alpha:FG\Rightarrow id_A$ and $\beta:GF\Rightarrow id_B$.

So, this would give the notion of Equivalence that I was looking for. This would give equivalence of categories, with no smooth structure involved.

Orbifolds as stacks? in page no $8$ says that, there is no analogous theorem for smooth functors between Lie groupoids, there are many fully faithful essentially surjective smooth functors between Lie groupoids with no continuous (I guess he mean smooth) weak inverse ($G:B\rightarrow A$ that I have mentioned above, he is calling it weak inverse).

Further it says,

Not every fully faithful and essentially surjective smooth functor between two Lie groupoids should be considered

an equivalnece of Lie groupoids(cf., not every smooth bijection between manifolds is a diffeomorphism).

He then says the accepted definition is what I have given above.

**Question $1$** : How does one see that the condition $2$ says the functor is faithful.

**Question $2$** : Yes, just declaring a smooth functor that is essentially surjective, fully faithful is not a good notion of equivalnece of Lie groupids. What is the motivation behind declaring the definition that I have given above is a good notion of equivalence of categories.

What is the extra thing we get if we declare morphisms between Lie groupoids to be not just equivalence of categories (fully faithful and surjective) but Morita equivalence of categories?

What is the extra thing we get if we...-- the next thing you wrote after that doesn't make sense to me. $\endgroup$29more comments