Why isn’t every vector bundle morphism an isomorphism

algebraic-topologydifferential-topologyvector-bundles

Let $(B,E,p)$ and $(B',E',p')$ be vector bundles with local trivializations $(U_\alpha,\phi_\alpha)$ and $(U'_\alpha,\phi'_\alpha)$. A vector bundle morphism is a pair $(F,f)$ so that

  • $f \circ p= p' \circ F $
  • $\forall b \in B: F:p^{-1}(b) \rightarrow p'^{-1}(f(b))$ is linear.

Now let $B=B'$. A vector bundle isomorphism is a pair $(F,f=id_B)$, so that

  • $(F,f=id_B)$ is morphism

  • $\forall b \in B: F:p^{-1}(b) \rightarrow p'^{-1}(b)$ is a vector space isomorphism.

According to the definition of a vector bundle
\begin{equation}
{ \varphi }_{ \alpha }{ | }_{ { p }^{ -1 }(b) }:{ p }^{ -1 }(b)\rightarrow \left\{ b \right\} \times { \mathbb{R} }^{ n }
\end{equation}
is a vector space isomorphism, leading to
\begin{equation}
{ p }^{ -1 }(b)\cong \left\{ b \right\} \times { \mathbb{R} }^{ n } \cong { p' }^{ -1 }(b)
\end{equation}
So the second requirement in the definition of a vector bundle isomorphism is already fulfilled and every morphism should be an isomorphism.

I don't think that's right, but I also don't understand where I'am wrong.

Edit: The two vector bundles have the same rank.

Best Answer

First of all, the two vector bundles could have different rank (in which case they could not possibly be isomorphic, but there could still be non-trivial maps between them), i.e., the $n$ of $p^{-1}(b) \cong \{b\} \times \mathbb{R}^n$ does not have to be the same as the $n'$ of $p'^{-1}(b) \cong \{b\} \times \mathbb{R}^{n'}$. And secondly, even if $n=n'$, not every map from $\mathbb{R}^n$ to $\mathbb{R}^n$ is an isomorphism. You are not just requiring that the fibers are isomorphic in some abstract way, but that the unique map induced by the morphism of vector bundles is an isomorphism. And this is far from being automatic. To make it more explicit, you have the following commutative diagram:

$$\require{AMScd} \begin{CD} p^{-1}(\{b\}) @>{F}>> p'^{-1}(\{b\})\\ @V{\varphi}VV @V{\varphi'}VV \\ \{b\} \times \mathbb{R}^n @>{\varphi' \circ F \circ \varphi^{-1}}>> \{b\} \times \mathbb{R}^{n'} \end{CD}$$

and $F$ is an isomorphism if and only if the bottom map is an isomorphism, but this is not automatic from the fact that $\varphi$ and $\varphi'$ are isomorphisms (where $\varphi:= \varphi_\alpha\vert_{p^{-1}(\{b\})}$ for some chart $U_\alpha$ containing $p^{-1}(\{b\})$, and similarly for $\varphi'$).

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