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.

  • 1
    $\begingroup$ do your bundles have to be of the same rank? $\endgroup$
    – janmarqz
    Commented Apr 8, 2020 at 20:29
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    $\begingroup$ Whenever you want to check some property of vector bundles, it's often useful to check it first for trivial bundles. Suppose we have two trivial bundles $E=B \times \mathbb{R}^n$ and $E'=B \times \mathbb{R}^{n'}$. Then it's easy to see that a morphism $E \to E'$ of vector bundles is the same as a linear map $\mathbb{R}^n \to \mathbb{R}^{n'}$. For nontrivial bundles, it's not true that a vector bundle morphism is a collection of linear maps for each fiber, since the bundles might have some nontrivial "twist" $\endgroup$
    – Exit path
    Commented Apr 8, 2020 at 20:39

1 Answer 1


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 $\varphi'$).

  • $\begingroup$ So $p^{-1}(b)$ is isomorphic to $p'^{-1}(b)$ via $\phi_\alpha$ but that doesn't matter because $F$ has to induce this isomorphism. $\endgroup$
    – NicAG
    Commented Apr 8, 2020 at 20:55
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    $\begingroup$ $F$ does not have to induce this isomorphism, it has to induce an isomorphism. I've added an edit, see if it's clearer now. The fact that $\varphi$ and $\varphi'$ are isomorphisms simply makes sure that the vector space structure on the fiber is well-defined and does not depend on the chart that you're considering. For $F$ to be an isomorphism is a completely different condition. $\endgroup$
    – 57Jimmy
    Commented Apr 9, 2020 at 8:59
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    $\begingroup$ If you have a map $F$ between vector bundles of different ranks, the fibers cannot be isomorphic in any way (and hence $F$ cannot be an isomorphism). If they have the same rank, the fibers are always isomorphic in some way (just because they have the same dimension as vector spaces, and vector spaces of the same dimension are always isomorphic), but the condition that $F$ induces an isomorphism is still far from being automatic, since not all maps between vector spaces of the same dimension are isomorphisms. $\endgroup$
    – 57Jimmy
    Commented Apr 9, 2020 at 9:06

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