I am currently working on one part of a problem surrounding sections and submanifolds.
Given a real vector bundle $\pi: E\rightarrow M$ of rank k, with a smooth global section $s:M\rightarrow E$, can the section be considered a submanifold of $E$ under its graph parametrization? Namely, for an open set $U\subseteq M$, consider $\Gamma(s)=\{(p,v)\in M\times \mathbb{R}^k|p\in U, v=s(p)\}$, which is a properly embedded m-dimensional submanifold of the product space $M\times \mathbb{R}^k$.
But the vector bundle itself has a local trivialization $h:\pi^{-1}(U)\rightarrow U\times \mathbb{R}^k$ which is a diffeomorphism from its open sets ($\pi^{-1}(U)$) to open sets in the product space($U\times \mathbb{R}^k$). Can one piece together a submanifold of $E$ using these local trivializations?
I am asking all this because, the problem I am working on involves showing that for a compact manifold, where the rank of the bundle is greater than the dimension of the space, has a non-vanishing section. Since the bundle $E$ has dimension $m+k$, and the zero section is a submanifold of dimension $m$, my plan was to try and use transversality to show the two sections can not intersect due to dimensional considerations. Also unclear on where I might use compactness so I'm still struggling.
Best Answer
Yes. The image of a (sufficiently smooth) section is generally a submanifold. But it might not be nice. Consider the trivial line bundle over the reals, which I'll visualize as $\mathbb R^2$, with projection $(x, y) \mapsto x$. Then a section really is just the graph of a function on the reals. When you think about things like $y = |x|$ and $y = x^3$ and $y = x^{3/2}$, you realize how awkward things can be. You clearly need some smoothness conditions.
Your proof sounds as if it's headed in exactly the right direction. There's a small difficulty: if you start wtih the zero-section, $z$, and another section, $s$, and make them transverse by adjusting $s$, will the perturbed $s$ still be a section? The perturbation might move it so that more than one point of the image of $s$ happens to lie in a single fiber. Back to the trivial $R^2 \to R$ example: suppose we have the section $s(x) = x^{(1/3)}$, i.e., $x = y^3$. You can perturb that to $x = y^3 - \epsilon y$, and now it's no longer a section.
So you need to apply transversality to a fiber-by-fiber perturbation; that may take a little more work (or not...I haven't done this in 30 years!)