Continuity/differentiability of parametrized pullback of forms Let $M^n$ be a differentiable manifold and $\omega$ be a $k$-form on $M$. Suppose we have a continuous (differentiable) function $F: M \times M \rightarrow M$, which we represent as a family of functions $(f_m)_{m \in M}$.
I am trying to see that the function $G: M \times M \rightarrow \Lambda^k(T^* M)$, $G(m, p) = (f_m^* \omega)_p$ is continuous (differentiable if $F$ also is).
While this seems intuitively obvious, I can't see a way to rigorously show this. Using trivializations of $\Lambda^k(T^* M)$ as $U \times \mathbb{R}^{\begin{pmatrix} n \\ k \end{pmatrix}}$ with $U$ open in $M$ seems absolutely hopeless, I cannot see a way to say something about $G^{-1}(U \times V)$ with $U \times V$ a basis set.
Although I framed this in a general context, my motivation for this comes from the case when $M$ is a Lie group and $F$ is the multiplication function.
As a side curiosity, what is the relationship between the continuity of the $G$ defined above and the map $m \in M \mapsto f_m^* \omega \in \Omega^k(M)$ with say the compact-open topology on $\Omega^k(M)$?
 A: I'll do the case where $M$ is a smooth manifold. Let $f:M\times M\to M$ be a smooth map. For each $p\in M$ let $f_p:M\to M$ be the smooth map given by $f_p(q)=f(p,q)$. For a differential $k$-form $\omega$ on $M$, let $g:M\times M\to\Lambda^kT^*M$ be the map given by $g(p,q)=(f_p^*\omega)_q$.
Let $V$ be a smooth vector field on $M$, then $h(p,q)=V_p(f_q)$ is smooth. We can see this by choosing local coordinates $x_i$ and a chart $\varphi$. Then $$h(p,q)=v_i(p)\left.\frac{\partial}{\partial x_i}\right\rvert_{p}(f_q)=v_i(p)\left.\frac{\partial}{\partial x_i}\right\rvert_{\varphi(p)}(f_q\circ \varphi^{-1})$$ which is obviously smooth.
Now let $u:M\times M\to \mathbb{R}$ be a smooth map. Then $h(p,q)=(du_p)_q$ is smooth. We can see this by letting $V$ be an arbitrary vector field on $M$. Then $(p,q)\mapsto h(p,q)(V_q)=(du_p)_q(V_q)=V_q(u_p)$ is smooth by the reasoning above. It follows that $h$ is smooth.
Let $u$ be as above, and let $\omega=du_p$, then $g$ is smooth. Again this follows from $g(p,q)(v)=(f_p^*\omega)_q(v)=(f_p^*du_p)_q(v)=(f_p^*du_p)_q(v)=du_p((df_p)_q(v))=d(u_p\circ f_p)_q(v)$. Which shows that $g(p,q)=d(u_p\circ f_p)_q$. Letting $u'(p,q)=u_p(f_p(q))$ in the above paragraph gives that $g$ is smooth for this $\omega$.
Now the statement follows for general $\omega$ by choosing local coordinates $x_i$ and linearity. So that $g(p,q)=(f_p^*\omega)_q=(f_p^*udx_{i_1}\wedge...\wedge dx_{i_k})_q=u\circ f_p(q)(f_p^*dx_{i_1})_q\wedge ...\wedge (f_p^*dx_{i_k})_q$.
