Is every vector bundle over a noncompact finite-dimensional manifold a summand of a trivial bundle? In the notes of Vector Bundles and K-theory by Prof Allen Hatcher, on page 12 he proved a Proposition that for each vector bundle $E\to B$ with $B$ compact hausdorff there exists a vector bundle $E'\to B$ such that $E\oplus E'$ is the trivial bundle.
Later in example 3.6 he showed that the compactness of $B$ is an important condition, otherwise canonical line bundle over $\mathbb{RP^\infty}$ would be a counter example.
My doubt is...  Is this Proposition still valid for finite dimensional manifold??? Or can someone provide me a counter example of a finite manifold where it is not true.
 A: I think that the proposition still holds for smooth finite-dimensional manifolds. Here is a sketch of an argument.
First of all, for a bundle of rank $k$ the following statements are equivalent.


*

*there exists a complementary bundle of rank $\ell$,

*the classifying map $X\to BGL_k(\Bbbk)=G_k(\Bbbk^\infty)$ factors through $G_k(\Bbbk^{k+\ell})$.


Next, consider an $n$-dimensional smooth manifold, then $M$ admits a triangulation. The stars of its barycentric subdivision provide an open cover $\{U_0,\dots,U_n\}$ of $M$. Here each $U_i$ is the disjoint union of the stars of those vertices that come from the $i$-simplices of the original triangulation. In particular, each $U_i$ is a disjoint union of contractible open subsets of $M$. Fix trivialisations of $E|_{U_i}$ for each $i$.
Then the transition maps $g_{ij}\colon U_i\cap U_j\to GL_k(\Bbbk)$ can be used to construct an explicit classifying map from $M$ to the $n$-fold join $(GL_k(\Bbbk)*\cdots*GL_k(\Bbbk))/GL_k(\Bbbk)$ in Milnor's construction of a classifying space.
Now, there is a homotopy equivalence
$$\operatorname{colim}_n(\underbrace{GL_k(\Bbbk)*\cdots*GL_k(\Bbbk)}_{n+1\text {factors}})/GL_k(\Bbbk)
\to\operatorname{colim}_\ell G_k(\Bbbk^{k+\ell})$$
between these two models for $BGL_k(\Bbbk)$.
Because each of the joins on the left is a finite CW complex, the restriction ends up in a finite-dimensional Grassmannian. In particular, the classifying map $f$ above factors through one of them. Together with the observation above, that proves the claim.
A: The proposition you refer to holds for any space homotopy equivalent to a finite dimensional CW complex. Here are the main points.


*

*The property that any vector bundle $E$ over a space has a complementary bundle $E^\prime$ is preserved by homotopy equivalences.

*Any finite dimensional CW complex is homotopy equivalent to a smooth manifold. (A classical result of Whitehead shows that any finite dimensional CW complex is homotopy equivalent to a locally finite finite dimensional CW complex. Then simplicial approximation gives a homotopy equivalent finite dimensional simplicial complex. Embed it into a Euclidean space, take a regular neighborhood, and smooth it.)

*The total space of any vector bundle $E$ over a smooth manifold $M$ smoothly embeds into a Euclidean space. The normal bundle  bundle of $E$ restricted to $M$ is the desired complementary bunlde $E^\prime$.  

*Note that any topological manifold is homotopy equivalent to a finite dimensional CW complex.
