Are Functions Smooth Sections?

Question

In differential geometry we often identify $\Omega^0(M)$ with the smooth functions on a smooth manifold $M$. But for every $i>0$ we know that: $$\Omega^i(M)=\Gamma(\Lambda^i(T^*M))$$ I imagine with $i=0$ then $\Lambda^i(T^*M)$ would be the trivial line bundle since constant functions exist on $M$. So can we think about functions as smooth sections of this trivial line bundle? If this is true, does that mean for every smooth function $f$ there is a projection such that: $$\pi\circ f=\text{Id}_M$$ If so, it would be very helpful to a problem I am currently working on...

Answer 1

Ok never mind, this is easy. Let $M\times \mathbb{R}$ be the trivial line bundle, then for any smooth function $f$ on $M$, we can easily obtain a map: $$\begin{align} M&\longrightarrow M\times\mathbb{R}\\ p&\longmapsto (p,f(p)) \end{align}$$ The projection onto $M$ denoted by $\pi_M$ then satisfies the above requirement.