Fix $n\in\mathbb{N}$. A vector bundle of rank $n$ is a smooth map $\pi:E\rightarrow B$ between manifolds such that $\forall p\in B: E_p := \pi^{-1}(p)$ is an $n$-dimensional vector space and $\forall p\in B$, there exists a neighborhood $U$ of $p$ and a diffeomorphism $\psi:E\mid_U:=\pi^{-1}(U)\rightarrow U\times \mathbb{R}^n$ such that $\operatorname{pr}_1\circ\psi = \pi$ and $\psi\mid_{E_q}:E_q\rightarrow\{q\}\times\mathbb{R}^n$ is a vectorspace isomorphism for all $q\in U$.
Let $\pi: E \rightarrow M$ be a vector bundle and let $s: M \rightarrow E$ be a smooth section. Show that $Im(s)$, the image of $s$, is a submanifold of $E$.
I have tried this exercise as follows. We know that the graph of $s$, $\Gamma(s) = \{(q,s(q) \, | \ q \in M \}$ is a submanifold of $M \times E$. Next, I tried to link $E$ to $\mathbb{R}^n$ via the fact that $E_p$ is an $n$-dimensional vector space and was planning then to apply the inverse of the diffeomorphism $\psi$ to obtain a submanifold in $M$.
As you can see, I am still missing a lot of details. I cannot figure it out, so I was asking myself if I was doing this exercise right.
Thanks in advance!
Edit: This is what I got from Clara Beschova's answer.
Let $p \in M$. Then there exists a local trivilization $\psi: \left.E\right|_U \rightarrow U \times \mathbb{R}$. We can then compose our section $s$ with $\psi$ to obtain a map $\left.s\right|_U: M \rightarrow \left.E\right|_U$ of the form $u \mapsto (u,f(u))$, with $f: U \rightarrow \mathbb{R}^n$ (It is of that form since $ s|_U$ is still a section and thus composition with the coordinate map must still give the identity). Then is indeed $Im(s) \cap E|_U = Im(s|_U) = gr(f)$ a submanifold of $U \times \mathbb{R}^n$.
My question: This open set $U$ and the map $f$ depend on the point $p \in M$. Because we have to prove that $Im(s)$ is a submanifold, we have to take the union of of all the submanifolds $gr(f)$ (which depend on the point $p$). Why is this still a submanifold? Thanks!
You could use that for a smooth map $f:X\to Y$ between smooth manifolds $f(X)\subseteq Y$ is an embedded submanifold if $f$ is an immersion which is also a homeomorphism onto its image. This can be derived e.g. from the constant rank theorem.
Here $\pi\circ s=\text{id}_M$, so $s$ is a homeomorphism onto its image. Since $d\pi\circ ds=d(\pi\circ s)=d(\text{id}_M)=\text{id}_{TM}$, $s$ is an immersion. Hence $Im(s)\subseteq E$ is an embedded submanifold.