Immersions of vector bundles Let $E$ be a vector bundle of rank $k$ over a smooth manifold $M$ of dimension $m$.  Then $E$ can be thought of as a $(m+k)$-dimensional manifold, so the (weak) Whitney immersion theorem states that $E$ can be immersed in $\mathbb{R}^{2(m+k)}$.  

Is it possible to immerse $E$ in $\mathbb{R}^{2m + k}$?  

 A: Here is an argument filling in the sketch provided by @MikeMiller in the comments.
Fix an immersion $M \looparrowright \mathbb{R}^{2m}$ with normal bundle $\nu_M$, which exists by the weak Whitney immersion theorem.  It suffices to show that there is a bundle monomorphism $E \hookrightarrow \nu_M \oplus \mathbb{R}^k$.  The tubular neighborhood theorem asserts that there is a diffeomorphism of $\nu_M$ onto a subspace of $\mathbb{R}^{2m}$, and so this gives an immersion of $E$ into $\mathbb{R}^{2m} \times \mathbb{R}^k \cong \mathbb{R}^{2m + k}$.  
It remains to show that such a bundle monomorphism exists.  This follows from the more general proposition.

Proposition.  Given a bundle $E$ of rank $k$ and a bundle $G$ of rank $k + m$ on a $m$-manifold $M$, there exists a bundle $F$ of rank $m$ such that $E \oplus F \cong G$.  

The proof of this proposition proceeds by obstruction theory.  The given bundles $E$ and $G$ furnish a classifying map $M \to BO(k) \times BO(k+m)$.  In terms of classifying maps, the conclusion of the proposition is equivalent to showing that the map $M \to BO(k) \times BO(k+m)$ lifts to $BO(k) \times BO(m)$:
$$
\require{AMScd}
\begin{CD}
@. BO(k) \times BO(m) \\
{} @V{1 \times \mu}VV \\ 
M @>>> BO(k) \times BO(k+m)
\end{CD}$$
Note that the cofiber of the vertical map has no positive-dimensional cells in dimensions $\leq m$, while $M$ can be approximated by a $m$-dimensional cell complex.  So the map from $M$ to the cofiber is null, and hence a lift exists.  
