I'm actually discovering Lie groupoids, and in the first place I took by definition that the Lie Groupoid inversion be smooth, but it came to my mind that it is a thing to demonstrate.
I tried to understand the demonstration in the Mackenzie K.C.H General Theory of Lie Groupoids but I failed to.
More precisely, this is the part that I don't understand:
Let $G$ the set of arrows, $G_0$ the set of objects of our groupoid and $\alpha$ and $\beta$ the source and target map respectively. Let $\theta : G \star G \longrightarrow G \times_\beta G$ a map defined by $(h,g) \longmapsto (h,hg)$ where $G \star G$ is the set of composable arrows and $G \times_\beta G$ the set of arrows that have the same target (and the product $hg$ is defined by the groupoid law.
$\theta$ is obviously bijective. I have a problem to show that it is an immersion: Let $(h,g) \in G \star G$ and $(Y,X) \in T_{(h,g)} G \star G$ and suppose that $T_{(h,g)}\theta (Y,X) = (0,0)$.
In this part the book states that $\pi_2 \circ \theta = \pi_2$ and that implies that $Y=0$. But $\pi_2$ is not defined, I think that it is the second projection. However, later in the proof they said that $\pi_2 : G \star G \longrightarrow G$, but how can we then compose $\pi_2 \circ \theta$, that does not make any sens for me.
In addition, if we admit that $\pi_2$ is really the second projection defined in $G \times G$, $\pi_2 \circ \theta$ is equal to the law of the groupoid.
I'd really appreciate your help, thanks a lot.