Let $G$ be a Lie group acting smoothly on a smooth manifold $M$. Suppose that the orbit space $M / G$ is a topological manifold, and is endowed with a smooth structure such that:


1. the quotient map $\pi : M \to M / G$ is smooth, and
2. the pullback map $\pi^\* : C^\infty(M / G) \to C^\infty(M)^G$ is an isomorphism, where $C^\infty(M)^G$ is the space of $G$-invariant smooth functions on $M$.


Is $\pi$ necessarily a submersion?



Quotient map is a submersion
**Edit**: This answer is wrong, I was misled by the complex case. The map $\pi$ is not surjective, its image is a semi-algebraic set. In fact it would seem that $M/G$ is never a manifold.


No. Take $M=\mathbb{R}^2$, $G=\mathbb{Z}/2$ acting by swapping the coordinates. The quotient is isomorphic to $\mathbb{R}^2$, with $\pi :\mathbb{R}^2\rightarrow \mathbb{R}^2$ given by $\pi (x,y)=(x+y,xy)$. By a general result of G. Schwarz, $\pi^\*:C^{\infty}(\mathbb{R}^2)\rightarrow C^{\infty}(\mathbb{R}^2)^G$ is an isomorphism. But $\pi $ is not a submersion along the diagonal $\{(x,x)\} $.


