# Smooth map between vector bundles from local coordinates

For a map $f: M \rightarrow N$ between smooth manifolds, the general strategy to show smoothness is to use

Lemma: $f: M \rightarrow N$ is smooth iff for every $x \in M$ there is a chart $(U, \Psi)$ at $x$ in $M$, and a chart $(V, \varphi)$ at $f(x)$ in $N$ such that the map $\Psi^{-1} \circ f \circ \varphi: \mathbb{R}^{dim(M)} \rightarrow \mathbb{R}^{dim(N)}$ is smooth.

So, we look for local coordinate representations of the function $f$ and show that this map is smooth.

How do we recreate this for vector bundles? Vector bundles are defined to be (smooth) manifolds, so it seems natural to try and use the above lemma. However, I'm having trouble unravelling the definitions of local charts on a vector bundle, so I can't even begin to obtain a coordinate representation of any function. My best guess is to use local trivialisations and somehow compose them with the local charts of the base manifolds, but I don't know how to continue.

I am correct in trying to use this lemma, or is there some easier strategy for working with maps between vector bundles?

• You are absolutely fine doing it this way. If you do it once or twice, or in my case about five hundred times, then this "first principles" method becomes appealing. – астон вілла олоф мэллбэрг Feb 4 '18 at 12:10
• How do I obtain charts in a vector bundle from local trivialisations? Is it always the case that a chart of $M$ induces a corresponding chart on $E$? – Doc Feb 4 '18 at 14:37
• Do you have any concrete examples of maps between vector bundles that you're interested in? – Kevin H. Lin Feb 5 '18 at 6:23
• Yes, the case where $E \oplus E^\bot \cong M \times \mathbb{R}^k$. I have the map $f: E \oplus E^\bot \rightarrow M \times \mathbb{R}^k$ defined fibrewise as $(v,u) \mapsto v+u$, and shown that it is fibrewise bijective, but I cannot prove that this is smooth. It seems like an underlying problem I am having to show smoothness of maps between vector bundles. I can do this OK for manifolds, but I don't even know what charts of a vector bundle look like – Doc Feb 5 '18 at 8:44
• It's "obviously" smooth -- the map is the identity "downstairs" and it's addition "upstairs". On a local trivialization it looks like $(+, \text{id})$. – Kevin H. Lin Mar 26 '18 at 23:33