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.

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    $\begingroup$ 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. $\endgroup$ – David Roberts Feb 7 at 0:02
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    $\begingroup$ 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. $\endgroup$ – David Roberts Feb 7 at 1:14
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    $\begingroup$ 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. $\endgroup$ – 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.

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    $\begingroup$ Nice! I didn't think of using essential surjectivity to get the object component to be a submersion. $\endgroup$ – David Roberts Feb 10 at 1:59

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