I've been working on solving the following problem:
Let $G$ be a connected Lie group acting (on the left) smoothly on the smooth oriented $m$-manifold $M$ via the map $\star : G\times M \to M$. Show that for each $g\in G$, if $\Phi_g : M \to M$ is the map $\Phi_g(p) = g\star p$, then $\Phi_g$ preserves the orientation on $M$.
My approach to this problem hinges on one fact that I'm having trouble properly justifying. Take a positively-oriented nonvanishing volume form $\omega \in \Omega^m(M)$. Then, for fixed $p\in M$ and a given oriented basis $(\mathbf{v}_1,\cdots,\mathbf{v}_m)$ of $T_pM$, consider the function $f : G\to \mathbb{R}$ given by $$ f(g) = \left[\Phi_g^*(\omega)\right]_p(\mathbf{v}_1,\cdots,\mathbf{v}_m) = \omega_{\Phi_g(p)}\left(d(\Phi_g)_p(\mathbf{v}_1), \cdots, d(\Phi_g)_p(\mathbf{v}_m)\right). $$ Once I can show $f$ is continuous, the rest of my argument will go through. Question: How can I show $f$ is continuous?
I know intuitively that $f$ should be smooth even, because $\omega$ and the action $\star$ are such. Additionally, I know from this answer that $f$ is related to the pullback form $\star^*(\omega)$, namely, $$ f(g) = \left[\star^*(\omega)\right]_{(g,p)}\left((\mathbf{0}_g,\mathbf{v}_1),\cdots, (\mathbf{0}_g,\mathbf{v}_m)\right) $$ because $d\star_{(g,p)}(\mathbf{0}_g,\mathbf{v}_j) = d(\Phi_g)_p(\mathbf{v}_j)$ for each $j$. Therefore, the problem reduces to arguing why $g\mapsto \left[\star^*(\omega)\right]_{(g,p)}$ "varies continuously" in $g$. This is what I'm having trouble understanding.
Perhaps the most concrete answer I'm looking for is this: if $g_0\in G$ is such that $f(g_0) > 0$, I want a description of the open set $U\subseteq G$ containing $g_0$ such that $f(U) \subseteq (0,\infty)$.