Requirement for weak pullback to be a Lie groupoid (Moerdijk)

Let $$\phi:\mathcal{G}\rightarrow \mathcal{K}$$ and $$\psi:\mathcal{H}\rightarrow \mathcal{K}$$ be morphisms of Lie groupoids.

We define weak pullback/2-fibre product corresponding to $$\phi:\mathcal{G}\rightarrow \mathcal{K}$$ and $$\psi:\mathcal{H}\rightarrow \mathcal{K}$$ to be a groupoid $$\mathcal{G}\times_{\mathcal{K}}\mathcal{H}$$ whose object set is $$(\mathcal{G}\times_{\mathcal{K}}\mathcal{H})_0=\{(a,\alpha,b)| a\in \mathcal{G}_0, b\in \mathcal{H}_0,\alpha:\phi(a)\rightarrow \psi(b) \in \mathcal{K}_1\}.$$ Given $$(a,\alpha,b ),(a',\alpha',b')\in (\mathcal{G}\times_{\mathcal{K}}\mathcal{H})_0$$ we declare $$\text{Mor}((a,\alpha,b),(a',\alpha,b'))=\{u:a\rightarrow a'\in \mathcal{G}_1, v:b\rightarrow b'\in \mathcal{H}_1| \alpha'\circ \phi(u)=\psi(v)\circ \alpha\}.$$

We then see that $$(\mathcal{G}\times_{\mathcal{K}}\mathcal{H})_0= \mathcal{G}_0\times_{\phi,\mathcal{K}_0,s}\mathcal{K}_1\times_{t\circ pr_2,\mathcal{K}_0,\psi}\mathcal{H}_0$$

$$(\mathcal{G}\times_{\mathcal{K}}\mathcal{H})_1=\mathcal{G}_1\times_{t\circ \phi,\mathcal{K}_0,s}\mathcal{K}_1\times_{t,\mathcal{K}_0,s\circ \psi}\mathcal{H}_1.$$ Source, target maps are given by $$s(u,\gamma,v)=(s(u),\gamma\circ \phi(u),s(v))$$ $$t(u,\gamma,v)=(t(u),\psi(v)\circ \gamma, t(v))$$

Moerdijk (in page no $$5$$) says that assuming $$t\circ pr_2:\mathcal{G}_0\times_{\mathcal{K}_0}\mathcal{K}_1\rightarrow \mathcal{K}_0$$ is a submersion confirm that $$\mathcal{G}\times_{\mathcal{K}}\mathcal{H}$$ is a Lie groupoid. All I can see is $$(\mathcal{G}\times_{\mathcal{K}}\mathcal{H})_0$$ and $$(\mathcal{G}\times_{\mathcal{K}}\mathcal{H})_1$$ are manifolds (being pullbacks under submersions) if $$t\circ pr_2:\mathcal{G}_0\times_{\mathcal{K}_0}\mathcal{K}_1\rightarrow \mathcal{K}_0$$ is a submersion.

But, what I do not understand is, why does $$s,t$$ are smooth?

For source map, first and third coordinates are submersions. Second projection $$\gamma\circ \phi(u)$$ does not seem to be submersion, unless I assume $$\phi:\mathcal{G}_1\rightarrow \mathcal{K}_1$$ is a submersion. Similarly, to prove target map is submersion, first and third projections are submersions but second projection is $$\psi(v)\circ \gamma$$ does not seem to be submersion unless I assume $$\psi:\mathcal{H}_1\rightarrow \mathcal{K}_1$$ is a submersion.

I think we should also assume $$\phi:\mathcal{G}_1\rightarrow \mathcal{K}_1$$ and $$\psi:\mathcal{G}_1\rightarrow \mathcal{K}_1$$ are submersions.

Does it follow with out assuming $$\phi,\psi$$ are submersions?

If $$\phi$$ is submersion, then, $$\mathcal{G}_1\times_{t\circ \phi,\mathcal{K}_0,s} \mathcal{K}_1\rightarrow \mathcal{K}_1\times_{t,\mathcal{K}_0,s}\mathcal{K}_1$$ given by $$(u,\gamma)\mapsto (\phi(u),\gamma)$$ is submersion. As multiplication map $$\mathcal{K}_1\times_{t,\mathcal{K}_0,s}\mathcal{K}_1\rightarrow \mathcal{K}_1$$ is submersion (thanks to David Roberts), the composition map given by $$(u,\gamma)\mapsto (\phi(u),\gamma)\mapsto \gamma\circ \phi(u)$$ is submersion.

So, for $$(u,\gamma,v)\mapsto \gamma\circ \phi(u)$$ to be submersion, we need $$\phi$$ to be submersion even after using multiplication map is submersion.

What am I missing here?

• The composition map in a finite-dimensional Lie groupoid is a submersion. I've not seen this anywhere, I had to prove it myself. – David Roberts Feb 15 '19 at 12:40
• Oh.. Thanks. I will check it :) @DavidRoberts – Praphulla Koushik Feb 15 '19 at 13:23
• @DavidRoberts I see why it is submersion... Given a Lie group $G$ the multiplication map $m:G\times G\rightarrow G$ is a submersion... Same idea for Lie groupoid also.. the multiplication map (composition) is submersion... so source, target maps for groupoid mentioned above are submersions... – Praphulla Koushik Feb 15 '19 at 15:48
• Allow me to consider a special case of this question: is it necessary for $M \to N \leftarrow M'$ to be submersions for $M\times_N M'$ to be a manifold? In fact no, since transversality is a sufficient condition. And, if $M$ is a closed submanifold of $N$, and $M'=M$, then $M\times_N M = M\cap M = M$ is a manifold, and the derivative of both inclusions is nowhere surjective. – David Roberts Mar 27 '19 at 21:37
• @DavidRoberts that’s true. For pullback to be manifold, all we need is intersection is transversal.. Here I am trying to see if $t\circ pr_2: \mathcal{G}_0\times_{\mathcal{K}_0}\mathcal{K}_1\rightarrow \mathcal{K}_0$ being submersion is sufficient to prove $\mathcal{G}\times_{\mathcal{K}}\mathcal{H}$ is a Lie groupoid.... – Praphulla Koushik Mar 29 '19 at 3:21

Lemma Given a diagram $$\array{ U & \to & V & \leftarrow &W \\ \downarrow && \downarrow && \downarrow \\ X & \to & Y & \leftarrow & Z }$$ of finite dimensional manifolds and submersions, then the induced map $$U\times_V W \to X\times_Y Z$$ is a submersion.
This is proved by taking $$(u,w) \in U\times_V W$$ and looking at the diagram of tangent spaces and seeing that the induced linear map at $$(u,w)$$ is surjective.
You can iterate this lemma to deal with the case of maps between limits of longer zig-zags, like $$A \to B \leftarrow C \to D \leftarrow E$$.