Let $f:X \to Y$ be continuous map between topological spaces and $E$ a $n$-vector bundle over $Y$. Then we know that the pullback $f^*E$ is also a $n$-bundle over $B$ and a pullback in sense of category theory.

Let assume that we have following commutative square

$$ \require{AMScd} \begin{CD} F @>{g} >> E \\ @VVpV @VVpV \\ X @>{f}>> Y \end{CD} $$

with another $n$-bundle $F$ and $g$ a bundle map. Since $f^*E$ is a pullback we obtain a continuous map $h:F \to f^*E$ of vector bundles.

My question is when is $h$ an isomorphism of vector bundles?

Does it suffice that $h$ induce a linear isomorphism $h_x: F_x \to f^*E_x$ on each fiber?

My considerations:

According to definition a morphism of vector bundles is an isomorphism iff

(1) it is a homeomorphism

(2) fiberwise a linear isomorphism

$h$ is by construction continuous and bijective. (2) holds by assumption.

Is then $h$ already an isomorphism or are there some pathological conterexamples? The problem is that I can't find an argument that the inverse map is also continuous.

Or does it suffice to verify that the map is local homeomorphism using trivialisations?

  • 1
    $\begingroup$ I think you must require that $g$ is bundle map. $\endgroup$
    – Paul Frost
    Mar 29, 2019 at 9:51
  • $\begingroup$ @PaulFrost: yes sorry, of course. $\endgroup$
    – user267839
    Mar 29, 2019 at 11:15

1 Answer 1


Claim: If $E\to X$, $E'\to X$ are two vector bundles over the same space $X$ and $f: E \to E'$ is a bundle map which is an isomorphism on each fiber, $f$ is an isomorphism, then $f$ is an isomorphism of bundles.

Proof: We know that $f$ is bijective; to show $f$ is an isomorphism, we need to show that it is a homeomorphism. As this is a local property, we may reduce to the case when $E $ and $E'$ are trivial bundles, i.e. we have a map $f: X \times \mathbb{R}^n \to X \times \mathbb{R}^n$ which is an isomorphism on each fiber. Then $f$ is of the form $(x,v) \mapsto (x, g(x)v)$ for some continuous map $g: X \to GL_n(\mathbb{R})$. Then the inverse map $\eta: GL_n(\mathbb{R}) \to GL_n(\mathbb{R})$, $\eta(A) = A^{-1}$ is continuous, so the map $(x,v) \mapsto (x,\eta(g(x)) v)$ is continuous, and this is an inverse for $f$.

In general, a vector bundle map $X \times \mathbb{R}^n \to Y \times \mathbb{R}^k$ is of the form $(x,v) \mapsto (f(x), g(x)v)$ for $f: X \to Y$ and $g: X \to \mathrm{Hom}(\mathbb{R}^n, \mathbb{R}^k)$, and any map of vector bundles looks like this map locally. This can be used to advantage.

  • $\begingroup$ Where did "$f$ is of the form $(x,v) \mapsto (x, g(x)v)$ for some continuous map $g: X \to GL_n(\mathbb{R})$." come from? $\endgroup$
    – user5826
    Dec 3, 2019 at 22:47
  • $\begingroup$ $f$ has to map the fiber over $x$ to the fiber over $x$, and it must be a linear isomorphism over each fiber, so it is of the form $(x,v) \mapsto (x,g(x)v)$. Continuity of $f$ implies that $g$ is continuous. $\endgroup$ Dec 8, 2019 at 14:46

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