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?

nowheresurjective. $\endgroup$ – David Roberts Mar 27 '19 at 21:37