Quotient of compact group action

Question

Suppose $K$ is a compact (say Lie) group acting on a manifold $N$ and we know that the orbit space $N/K$ is compact. Is there a simple reason why $N$ must be compact?

Thanks!

Answer 1

For every $x\in N$, there exists an open subset $U_x,$ such that the adherence $V_x$ of $U_x$ is compact. Let $p:X\rightarrow N/K$ be the quotient map. The family $(p(U_x))_{x\in N}$ is an open covering of $N/K$. Since $N/K$ is finite, we can extract from it a finite covering $p(U_1),...,p(U_n)$. Recall that the adherence $V_i$ of $U_i$ is compact. Let $f_i: V_i\times K\rightarrow N$ defined by $f_i(y,k)=k.y$, the image of $f_i $ is compact since $V_i\times K$ is compact, this implies that $V=\bigcup_if_i(V_i\times K)$ is compact since it is a finite Union of compact set. Let us show that $V$ is $N$. Let $z\in N$, there exists $i\in \{1,...,n\}$ such that $p(z)\in p(U_i)$. This implies that there exists $w\in U_i\subset V_i$ such that $p(w)=p(z)$. This is implies the existence of an element $k\in K$ such that $z=k.w$ and $z$ is in the image of $f_i$.