I'm going through Hirsch's "Differential Topology" and am trying to make my way through the proof of Theorem 5.3 on Page 112.
The theorem is Let $M \subset V$ be a sub manifold, $\delta M = \delta V = \emptyset$. Then any two tubular neighborhoods of $M$ in $V$ are isotopic.
Let $(f_i, \gamma_i)$ for $i=0,1$ be our tubuluar neighborhoods. We first prove the theorem under the assumption that $f_0(E_0) \subset f_1(E_1)$.
The text then reads:
"Let $\Phi: \gamma_0 \rightarrow \gamma_1$ be the fibre derivative of $g = f_1^{-1} f_0: E_0 \rightarrow E_1$. Thus $\Phi$ is the component along the fibres of the morphism:
$T_M g: T_M E_0 = TM \oplus \gamma_0 \rightarrow TM \oplus \gamma_1 = T_M E_1$
Which shows that $\Phi$ is an isomorphism of vector bundles.
What exaclty does it mean to say that "Thus $\Phi$ is the component along the fibres"? Can somebody reword this for me give me some insight, whether technical or intuitive.
Thanks!!
Hirsch defines what it is in that sentence. Here $E$ and $F$ are vector bundles over $M$ and you have a bundle map $\Phi\colon E\to F$ covering the identity map on $M$. If you restrict $d\Phi$ to the (tangent space to the) fiber $E_p$, this is the fiber derivative, mapping $E_p$ to $F_p$. Using a trivialization over $U\subset M$, you can write coordinates $(x,v)$ on $E|_U$ and $(x,w)$ on $F|_U$. (Fixing $x$ and varying $v$ over $\Bbb R^k$ represents the fiber of $E$ over $x$.) Then $\Phi$ looks like $$g(x,v) = (x,h(x,v))$$ and the fiber derivative at $(x,v)$ on the tangent vector $(0,\xi)$ to the fiber is given by $dh_{(x,v)}(0,\xi)$.