# Isomorphism of Vector Bundles

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?

• I think you must require that $g$ is bundle map. Mar 29, 2019 at 9:51
• @PaulFrost: yes sorry, of course. Mar 29, 2019 at 11:15

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.
• 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? Dec 3, 2019 at 22:47
• $f$ 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. Dec 8, 2019 at 14:46