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.

  • the square $$\require{AMScd}\begin{CD} \mathcal H_1 @>\phi>> \mathcal G_1 \\ @V(s, t)VV @V(s, t)VV \\ \mathcal H_0 \times \mathcal H_0 @>\phi \times \phi>> \mathcal G_0 \times \mathcal G_0 \end{CD}$$

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 the case of simple examples but did not succeed.

The 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 suggested by Benjamin Steinberg, I tried to see that the first condition is saying $\phi$ is an essentially surjective functor and the second condition is saying that $\phi$ is fully faithful.

I was able to see that the first condition implies that the functor is essentially surjective. I have deleted the proof here and added it in comments (1 2) so that the question does 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 an 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 equivalence 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 equivalence 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?

  • 1
    $\begingroup$ You should first check if you had discrete groupoids this would boil down to a full and faithful functor (that's what the second condition gives) Such that every object of G is isomorphic to one on the image (that's what the first condition says). Now Moerdijk is writing this diagramatically to build in the appropriate smoothness. $\endgroup$ May 24, 2018 at 17:02
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    $\begingroup$ It is just natural equivalence for discrete groupoids. Now you want to write fully faithful and essentially surjective in a diagramatic way so you can do a smooth version. Convince yourself first that the first condition is essentially surjective in the discrete case and the second is fully faithful. $\endgroup$ May 24, 2018 at 18:10
  • 3
    $\begingroup$ By the way, Morita equivalence of Lie groupoids can also be formulated in terms of principal bibundles that makes it look more like Morita contexts in ring theory. $\endgroup$ May 24, 2018 at 18:16
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    $\begingroup$ I don't think its a question of generalizing to the smooth manifold setting. What you have seen is Moerdijk's definition gives the natural definition for discrete groupoids so it is a reasonable definition in the smooth setting. $\endgroup$ May 25, 2018 at 9:55
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    $\begingroup$ Let $b\in G_0$. To prove the functor $\phi$ is essentially surjective, we need to prove that $b$ is isomrorphic to $\phi(a)$ for some $a\in H_0$. As $G_1\times_{G_0} H_0\xrightarrow{\pi_2}G_1\xrightarrow{t}G_0$ surjective, there exists an element $(g,h)\in G_1\times_{G_0}H_0$ such that $t(\pi_1(g,h))=b$. By $(g,h)\in G_1\times_{G_0}H_0$ we mean that $s(g)=\phi(h)$. $\endgroup$ Jun 26, 2018 at 14:15

1 Answer 1


(Using the same notations as mentioned in the question.)

Let $g \in \mathcal{G}_0$. Then the first condition ensures the existence of a $(\gamma , x) \in \mathcal{G}_1 \times{_{s,\mathcal{G}_0,\phi _0}} \mathcal{H}_0 $ such that $t(\gamma)=g$ and $s(\gamma)=\phi_0(x)$. Essential surjectivity follows from the fact that $\mathcal{G}$ is a groupoid.

The second condition ensures that there exists a diffeomorphism $\mathcal{F}:\mathcal{H}_1 \rightarrow \mathcal{G}_1 \times_{\mathcal{G}_0 \times \mathcal{G}_0} (\mathcal{H}_0 \times \mathcal{H}_0)$ given by $\gamma \mapsto \bigl( \phi_1(\gamma),(s(\gamma),t(\gamma) ) \bigr)$. The smooth structure on $\mathcal{G}_1 \times_{\mathcal{G}_0 \times \mathcal{G}_0} (\mathcal{H}_0 \times \mathcal{H}_0)$ is induced by the bijection
$$(\mathcal{G}_1 \times_{s,\mathcal{G}_0,\phi_0} \mathcal{H}_0)\times_{t \circ \pi_2, \mathcal{G}_0, \phi_0} \mathcal{H}_0 \rightarrow \mathcal{G}_1 \times_{(s,t), \mathcal{H}_0 \times \mathcal{H}_0, \phi \times \phi}(\mathcal{H}_0 \times \mathcal{H}_0)$$ given by $\bigl((\gamma,h_1),h_2) \bigr) \mapsto \bigl( \gamma, (h_1,h_2) \bigr)$.

Now, since $\operatorname{Hom}(h,k)$ is a closed submanifold of $\mathcal{H}_1$ for all $h,k \in \mathcal{H}_0$, hence $$\phi\rvert_{\rm{Hom}(h,k)}= \operatorname{pr}_1 \circ \mathcal{F}\rvert_{\operatorname{Hom}(h,k)}: \operatorname{Hom}(h,k) \rightarrow \operatorname{Hom}(\phi(h),\phi(k))$$ is a diffeomorphism for all $h,k \in \mathcal{H}_0$. Hence the second condition is not only saying $\phi\rvert_{\operatorname{Hom}(h,k)}$ is a bijection but in fact a diffeomorphism.

Note: The question was asked a long time back and it seems the issue has been resolved in the comments section but just for the sake of completeness for the future readers I added an answer here.

  • $\begingroup$ I did a spot of proofreading, and, while I was editing, noticed that all $\phi$s were \mathcald. Since this makes no difference (see $\phi$ \phi vs. $\mathcal{\phi}$ \mathcal{\phi}), I removed it, hopefully doing no harm to your intention. $\endgroup$
    – LSpice
    Jan 28, 2022 at 21:43
  • $\begingroup$ @LSpice Yh it's fine. $\endgroup$ Feb 17, 2022 at 15:03
  • $\begingroup$ @AdittyaChaudhuri Could you please elaborate on why $\mathrm{pr}_1\circ\mathcal F|_{\mathrm{Hom}(h,k)}$ should be a diffeomorphism? $\endgroup$
    – Žan Grad
    Dec 11, 2022 at 19:47
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    $\begingroup$ Oh, I think I see why: $\mathcal F|_{\mathrm{Hom}(h,k)}$ maps $\mathrm{Hom}(h,k)$ diffeomorphically onto $\mathcal H\times_{\mathcal G_0\times \mathcal G_0} \{(h,k)\}$ and the latter is identified with $\mathrm{Hom}(\phi(h),\phi(k))$. $\endgroup$
    – Žan Grad
    Dec 11, 2022 at 19:56

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