# Equivalence of definitions of equivalence of étale Lie groupoids

I've come across two definitions of an equivalence of étale Lie groupoids, and I'd like to know whether they are equivalent.

Let $$\mathcal{G}$$ be an étale Lie groupoid with space of objects $$\mathcal{G}_0$$ and space of arrows $$\mathcal{G}_1$$. Write $$\alpha$$ and $$\omega\colon \mathcal{G}_1 \to \mathcal{G}_0$$ for the maps sending an arrow to its source and target, respectively.

Definition 1. In Metric Spaces of Nonpositive Curvature, Bridson–Haefliger define an equivalence of étale (Lie) groupoids $$f\colon \mathcal{G} \to \mathcal{H}$$ to be a smooth functor such that $$f\colon \mathcal{G}_0 \to \mathcal{H}_0$$ is an étale map (a local diffeomorphism) and such that

1. for all $$x \in \mathcal{G}_0$$, the map $$f$$ induces an isomorphism of isotropy groups $$\mathcal{G}_x \to \mathcal{H}_{f(x)}$$, and
2. the map $$f$$ induces a bijection of orbit spaces $$\mathcal{G}_1\backslash\mathcal{G}_0 \to \mathcal{H}_1\backslash\mathcal{H}_0$$.

Definition 2. In "Orbifolds as Groupoids: An Introduction", Moerdijk defines an equivalence of (étale) Lie groupoids $$f\colon\mathcal{G} \to \mathcal{H}$$ to be a smooth functor such that

1. the map $$\omega\pi_1 \colon \mathcal{H}_1\times_{\mathcal{H}_0}\mathcal{G}_0 \to \mathcal{H}_0$$ defined on the fiber product $$\{(g,y) \in \mathcal{H}_1\times\mathcal{G}_0 : \alpha(g) = f(y)\}$$ sending a pair $$(g,y)$$ to $$\omega(g)$$ is a surjective submersion, and
2. the following diagram is a pullback square $$\require{AMScd}$$ $$\begin{CD} \mathcal{G}_1 @>f>> \mathcal{H}_1 \\ @VV{\alpha\times\omega}V @VV{\alpha\times\omega}V \\ \mathcal{G}_0\times\mathcal{G}_0 @>{f\times f}>> \mathcal{H}_0\times\mathcal{H}_0, \end{CD}$$ i.e. the map $$g \mapsto (\alpha(g),\omega(g),f(g))$$ is a diffeomorphism from $$\mathcal{G}_1$$ to $$\{ (x,y,g) \in \mathcal{G}_0\times\mathcal{G}_0\times\mathcal{H}_1 : \alpha(g) = f(x),\ \omega(g) = f(y)\}$$.

Remark. Moerdijk notes that the first condition says that each $$x \in \mathcal{H}_0$$ is isomorphic to some $$f(y)$$, so the map of orbit spaces is a surjection. He also notes that the second condition implies $$f$$ induces a diffeomorphism from the space of arrows $$g \colon x \to y$$ in $$\mathcal{G}$$ to the space of arrows $$g' \colon f(x) \to f(y)$$ in $$\mathcal{H}$$, so the map of orbit spaces is an injection, and for each $$x \in \mathcal{G}_0$$, the map $$f$$ induces an isomorphism of isotropy groups $$\mathcal{G}_x \to \mathcal{H}_{f(x)}$$. Thus Definition 2 implies all parts of Definition 1 except possibly the requirement that $$f$$ is an étale map.

• If $1\Rightarrow 2$, then it has to use something special about étale groupoids (as opposed to just being Lie groupoids). Definitely 1 seems weaker a priori. I think that the equivalence relation defined by 2 on étale Lie groupoids implies the one given by 1, even if for an individual functor 2 might not give 1. It certainly isn't true in general that a functor satisfying 2 satisfies 1. – David Roberts Feb 7 at 0:02
• Assuming $\mathcal{G}_1$ is a pullback as in the square, you can write it as an iterated pullback involving the source and target maps of $\mathcal{H}$ and the single map $f_0\colon \mathcal{G}_0\to \mathcal{H}_0$, namely the limit of $\mathcal{G}_0\to \mathcal{H}_0 \leftarrow \mathcal{H}_1 \to \mathcal{H}_0 \leftarrow \mathcal{G}_0$. By assumption the two maps $\mathcal{H}_1 \to \mathcal{H}_0$ are étale, and one can ask whether the maps $\mathcal{G}_1 \to \mathcal{G}_0$ being étale forces $f_0$ to be étale. It seems to me that assuming $f_0$ to be a submersion might force it. – David Roberts Feb 7 at 1:14
• But the definition only requires a given square to be a pullback, which only means that there is transversality, not that $f_0$ and hence $f_0\times f_0$ is a submersion. – David Roberts Feb 7 at 1:15

Notice that because $$\omega$$ is étale and $$\omega\pi_1$$ is a submersion, $$\pi_1\colon \mathcal{H}_1\times_{\mathcal{H}_0}\mathcal{G}_0 \to \mathcal{H}_1$$ is a submersion, as is $$\alpha\pi_1$$. Since $$\alpha$$ is étale, $$\pi_2\colon \mathcal{H}_1\times_{\mathcal{H}_0}\mathcal{G}_0 \to \mathcal{G}_0$$ is étale. Therefore because the composition $$\alpha\pi_1 = \pi_2f$$ is a submersion, so is $$f \colon \mathcal{G}_0 \to \mathcal{H}_0$$. Because $$f$$ induces a bijection of orbit spaces, only points in the same orbit may be identified by $$f$$. Since $$\mathcal{G}$$ is étale, its orbits are discrete. Therefore the dimension of $$\mathcal{G}_0$$ is equal to the dimension of $$\mathcal{H}_0$$ and we conclude that $$f\colon \mathcal{G}_0 \to \mathcal{H}_0$$ is étale. (Actually, $$f\colon \mathcal{G}_1 \to \mathcal{H}_1$$ is also étale). Therefore Definition 2 implies Definition 1.
Edit: Definition 1 implies Definition 2. Consider the fiber product $$\begin{CD} \mathcal{H}_1\times_{\mathcal{H}_0}\mathcal{G}_0 @>{\pi_2}>> \mathcal{G}_0 \\ @V{\pi_1}VV @VfVV \\ \mathcal{H}_1 @>\alpha>> \mathcal{H}_0. \end{CD}$$ Because $$f$$ is étale, $$\pi_1$$ is étale; therefore $$\omega\pi_1$$ is a fortiori a submersion. It is surjective as soon as, for every $$x \in \mathcal{H}_0$$, there exists $$y\in \mathcal{G}_0$$ and an arrow $$f(y) \to x$$. This is true because $$f$$ induces a bijection of orbit sets.
Now consider the fiber product $$\begin{CD} K = (\mathcal{G}_0\times\mathcal{G}_0)\times_{\mathcal{H}_0\times\mathcal{H}_0}\mathcal{H}_1 @>{\pi_2}>> \mathcal{H}_1 \\ @V{\pi_1}VV @V{\alpha\times\omega}VV \\ \mathcal{G}_0 \times\mathcal{G}_0 @>{f\times f}>> \mathcal{H}_0\times\mathcal{H}_0. \end{CD}$$
We want to show that the map $$F\colon \mathcal{G}_1 \to K$$ defined by $$g \mapsto (\alpha(g),\omega(g),f(g))$$ is a diffeomorphism. Notice that because $$f\times f$$ is étale and $$\pi_2F = f$$ is étale, $$F$$ is étale. We need therefore only show that it is a bijection. It is easy to argue that because $$f$$ induces a bijection of orbit sets and isomorphisms on isotropy groups, for each $$x$$ and $$y \in \mathcal{G}_0$$, the map $$f$$ induces a bijection from the set $$\{g\colon x \to y\}$$ to the set $$\{g' \colon f(x) \to f(y)\}$$. Therefore $$F$$ is a diffeomorphism.