# "Linear isomorphism" in definition of vector bundle

I'm reading out of Broecker & Jaenich's differential topology text, and in the definition of vector bundle I'm having trouble understanding what they're talking about. This trouble is worsened by the fact that other sources define the structure similarly, and also don't address my problem.

The definition is as follows:

"A ($n$-dimensional real topological) vector bundle is a trouble $(E, \pi X)$, where $\pi : E \to X$ is a continuous surjective map, every $E_x = \pi^{-1}(x)$ has the structure of an $n$-dimensional real vector space such that:

Axiom of local triviality. Every point of $X$ has a neighborhood $U$, for which there exists a homeomorphism $$f : \pi^{-1} (U) \to U \times \mathbb{R}^{n}$$ such that for every $x \in U$ $$f_{x} |E_{x} \to \{ x \} \times \mathbb{R}^{n}$$ is a vector space isomorphism." [Bold added for emphasis]

My trouble here is that we don't seem to have imposed any sort of vector space structure on $E_{x}$. Though the authors don't address exactly what kind of structures we've imposed on $E$ and $X$, other sources suggest only a topological space is needed, though others go on to imagine that $X$ is a topological or smooth manifold. None of these make reference to the assumption that any of the sets in question should have any kind of vector space structure. So I have no clue how I'm supposed to parse out what it means to have an isomorphism of vector spaces without a vector space. Is it supposed to be with reference to a chart on $X$? Someone please explain this.

Thanks.

• At the beginning of the definition it says "every $E_x = \pi^{-1}(x)$ has the structure of an $n$-dimensional real vector space"... Dec 21, 2016 at 4:32
• @EricWofsey I'd thought it was going on to formalize what that meant, i.e. it was saying $\pi^{-1}(x)$ has the structure of an $n$-dimensional real vector space, which was meant in the sense of the space satisfying the local triviality condition. So could I instead say that given a continuous surjection $\pi$, where each point $x \in X$ has a neighborhood $U$ such that $\pi^{_1}(U) \simeq U \times \mathbb{R}^{n}$, I can induce a vector space structure for $\pi^{-1}(x)$ by pulling back from $\{ x \} \times \mathbb{R}^{n}$?
– AJY
Dec 21, 2016 at 4:37
• No, those vector space structures are part of what it means to specify a vector bundle. See my answer. Dec 21, 2016 at 4:41

Part of the structure of a vector bundle $(E,\pi,X)$ is, as your definition says, a vector space structure on the set $E_x$ for each $x\in X$. That is, to have a vector bundle, you have to specify a vector space structure on each $E_x$. So the maps $f_x$ are supposed to be vector space isomorphisms with respect to this specified vector space structure on $E_x$ (and the obvious vector space structure on $\{x\}\times \mathbb{R}^n$).
(For a topological vector bundle, as in the definition you quoted, you also need topologies on $E$ and $X$, in order to say that $\pi$ is continuous and $f$ is a homeomorphism. For smooth vector bundles, you require $E$ and $X$ to be smooth manifolds and require all the maps to also be smooth.)