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Pullbacks of stacks coming from Lie groupoids are not always equivalent to Lie groupoids. Take G=H=R (the real line). Define F(x)=0 if x≤0 and F(x)=exp(−1/x^2) if x>0. The pullback is not equivalent to a Lie groupoid in this situation: the set-theoretical pullback is (−∞,0]⨯(−∞,0]∪{(x,x)|x∈R}, which is clearly not a smooth manifold.

One can guarantee that the pullback is a Lie groupoid by imposing transversality conditions on the maps involved. That is, A ⨯_C B is a Lie groupoid if A_0 → C_0 ← B_0 is transversal and A_1 → C_1 ← B_1 is transversal, where subscripts 0 and 1 denotes objects and morphisms respectively. (In fact, this transversality condition guarantees that the pullback is also a homotopy pullback, which is almost always what one actually wants.)

Pullbacks of stacks coming from Lie groupoids are not always equivalent to Lie groupoids.

Take $G=H=\mathbb{R}$. Define $F(x)=0$ if $x\leq 0$ and $F(x)=exp(−1/x^2)$ if $x>0$.

The pullback is not equivalent to a Lie groupoid in this situation: the set-theoretical pullback is $(−\infty,0]\times(−\infty,0]\cup \{(x,x)|x\in \mathbb{R}\}$, which is clearly not a smooth manifold.

One can guarantee that the pullback is a Lie groupoid by imposing transversality conditions on the maps involved. That is, $A \times_C B$ is a Lie groupoid if $A_0 \rightarrow C_0 \leftarrow B_0$ is transversal and $A_1 \rightarrow C_1 \leftarrow B_1$ is transversal, where subscripts $0$ and $1$ denotes objects and morphisms respectively. (In fact, this transversality condition guarantees that the pullback is also a homotopy pullback, which is almost always what one actually wants.)

Your first claim is false already for very simple cases. Take G=H=R (the real line). Define F(x)=0 if x≤0 and F(x)=exp(−1/x^2) if x>0. The pullback is not a Lie groupoid in this situation: the set-theoretical pullback is (−∞,0]⨯(−∞,0]∪{(x,x)|x∈R}, which is clearly not a smooth manifold.

Pullbacks of stacks coming from Lie groupoids are not always equivalent to Lie groupoids. Take G=H=R (the real line). Define F(x)=0 if x≤0 and F(x)=exp(−1/x^2) if x>0. The pullback is not equivalent to a Lie groupoid in this situation: the set-theoretical pullback is (−∞,0]⨯(−∞,0]∪{(x,x)|x∈R}, which is clearly not a smooth manifold.

One can guarantee that the pullback is a Lie groupoid by imposing transversality conditions on the maps involved. That is, A ⨯_C B is a Lie groupoid if A_0 → C_0 ← B_0 is transversal and A_1 → C_1 ← B_1 is transversal, where subscripts 0 and 1 denotes objects and morphisms respectively. (In fact, this transversality condition guarantees that the pullback is also a homotopy pullback, which is almost always what one actually wants.)

Your first claim is false already for very simple cases. Take $G=H=\mathbb R$ (the real line). Define $F(x)=0$ if $x≤0$ and $F(x)=\exp(−1/x^2)$ if $x>0$. The pullback does not exist in this situation: the set-theoretical pullback would have to be $(−\infty,0]⨯(−\infty,0]\cup\{(x,x)|x\in \mathbb R\}$, which is clearly not a smooth manifold.

Your first claim is false already for very simple cases. Take G=H=R (the real line). Define F(x)=0 if x≤0 and F(x)=exp(−1/x^2) if x>0. The pullback is not a Lie groupoid in this situation: the set-theoretical pullback is (−∞,0]⨯(−∞,0]∪{(x,x)|x∈R}, which is clearly not a smooth manifold.