In the proof of the Lefschetz formula for smooth maps $\phi\colon M \to M$ via the Heat Kernel (See Roe Elliptic Operators, Topology and asymptotic methods chapter 10) it's done the following observation:
Let $\phi \colon M \to M$ a smooth map ($M$ closed smooth oriented), let $(S,d)$ be a Dirac complex, let $\zeta \colon \phi^*S \to S$ a smooth bundle map, together they define a geometric endomorpism $$ F:=\zeta\phi^* \colon C^{\infty}(S)\to C^{\infty}(S)$$ the smooth sections of my bundle $S$.
Given now a smoothing operator $$P\colon L^2(S)\to C^{\infty}(S)$$ where $L^2(S)$ are the $L^2$-sections of $S$, it's claimed that $$FP\colon L^2(S)\to C^{\infty}(S)$$ is again a smoothing operator
What I tried: let $k\ \colon M\times M \to S\boxtimes S^*$ be the smooth kernel of $P$ (think of $k(m_1,m_2)\in \hom(S_{m_2},S_{m_1})$), then for any section $s\in L^2(S)$, we have $$ FP(s)(m_1)=F\int_M k(m_1,m_2)s(m_2)vol(m_2)$$
Now upon carefully looking at what $F=\zeta\phi^*$ does on a section $s\colon M \to M$, We see that $$ FP(s)(m_1)=\zeta P(s)(\phi(m_1))=\zeta\circ \int_M k(\phi(m_1),m_2)s(m_2)vol(m_2)$$
I would like to move my bundle map $\zeta$ inside the integral. If I'm allowed to do that I think I'm able to prove (following what's done by Roe at page $135$) that we still have a smooth kernel. Maybe using partition of unity (we can take a finite cover subordinate to the trivialising cover) we can reduce everything to (a finite sum of) integrals of vector valued functions over opens $U_i$, and then $\zeta$ would be a finite matrix whose entries depend on $m_1$) which should commute with the integral.
Can someone provide some feedback or ideas on how to do that? (Roe "does" it for the heat kernel at page 135 without any justification)
Some context (especially to understand more the notation) can be found here