Is the product of source and target maps of a Lie Groupoid a submersion? Let $G:= (G_1 \rightrightarrows G_0)$ be a Lie Groupoid. By definition, we know that source $s$ and target $t$ are submersion. Now define $(s,t):G_1 \rightarrow G_0 \times G_0$ as $\gamma \mapsto (s(\gamma),t(\gamma))$.
My Question is the following:
(1) Is $(s,t)$ a submersion?
(2) More generally, is product of two submersions always a submersion?
I am expecting a positive answer.
My attempt in the context of my original question:
$(s,t)_{*,\gamma}:T_\gamma(G_1) \rightarrow T_{(s(\gamma),t(\gamma))}(G_0 \times G_0) \cong T_{s(\gamma)}(G_0) \times T_{t(\gamma)}(G_0)$ (Standard identification) where $\gamma \in G_1$ and $(s,t)_{*. \gamma}$ is the differential of $(s,t)$ at $\gamma \in G_1$.
Also, $s_{*,\gamma} \times t_{*,\gamma} : T _{\gamma} (G_1) \rightarrow T_{s(\gamma)}(G_0) \times T_{t(\gamma)}(G_0)$ defined by $\lambda \mapsto (s_{* ,\gamma}(\lambda), t_{* ,\gamma}(\lambda))$
Now I am guessing that $(s,t)_{*,\gamma} = s_{*,\gamma} \times t_{*,\gamma}$ ....(3)
But I am not able to prove it explicitly.
More generally, is there any result along the following line:
If $F:M \rightarrow N \times N$ is smooth map of finite dimensional smooth manifolds given by $F(x)=(f(x) , g(x))$ where $f ,g :M \rightarrow N$ are smooth maps. Then is $F_{*,p} = f_{{*,p}} \times g_{{*,p}} $?... (4)
After this I am not able to proceed!
Summarising I asked Questions marked (1) ,(2) ,(3) ,(4) with an ultimate goal to get the answer of (1) that is
Is $(s,t)$ a submersion?
Thanks in advance.

In general, the map $(s,t):G_1\rightarrow G_0\times G_0$ need not be a submersion.
Suppose $M$ is a positive dimensional manifold and consider the Lie groupoid $M \rightrightarrows M$ where we think of the objects as being $M$, and for each $m\in M$, the only arrow at $m$ is the identity arrow (so the collection of arrows is canonically identified with $M$).
Then $s = t = Id_M$.  So, $(s,t):M\rightarrow M\times M$ is the diagonal map $m\mapsto (m,m)$.  The dimension of $M\times M$ is twice the dimension of $M$, so for positive dimensional $M$, there can be no submersion $M\rightarrow M\times M$.  Thus, $(s,t)$ is not a submersion in this case.
