# Why does every vector bundle admit a map into the universal bundle?

The following question regards a proof about Lemma 5.3 in Milnor-Stasheff. The proposition is to prove that every rank $$n$$ vector bundle $$\pi:E\to B$$ with compact (more generally, paracompact) base admits a map to the universal bundle over $$G_n$$, where $$G_n$$ is the set of n-planes in $$\mathbb{R}^L$$ for some large $$L$$.

This is done by showing that $$E$$ admits a linear, injective map on each fiber into $$\mathbb{R}^L$$ (injectivity implies we don't collapse any n-plane). My question concerns the injectivity of the map we eventually construct.

We choose a finite cover of the base $$U_i$$ where we may trivialize the bundle over each $$U_i$$. From here, the proof uses two ingredients. For each $$i$$,

1. Consider the map $$h_i:\pi^{-1}(U_i)\to \mathbb{R}^n$$ by trivialization composed with the projection.
2. Consider $$U_i\supset V_i\supset W_i$$, and construct a continuous cutoff function $$\lambda_i:B\to \mathbb{R}$$ with respect to these inclusions.

We combine these two ingredients together to form the map $$h_i':E\to \mathbb{R}^n$$, where

$$h_i':=\begin{cases} \lambda_i(\pi(e))\cdot h_i(e)&\pi(e)\in U_i\\ 0&\pi(e)\not \in V_i \end{cases}$$

and we claim the map $$e\mapsto (h_1'(e),...,h_r'(e))$$ is the injective, linear map. Injectivity is assured iff each nonzero $$e$$ has some nonzero component. However, I do not see why this needs to be the case. Let's take $$V_i$$ extremely small, so that each $$V_i\cap V_j=0$$ for $$i\neq j$$. In this case, if $$\pi(e)\in U_i-V_i$$ with nonzero $$e$$, then $$h_j'(e)=0$$ for each $$j$$. To me it seems like we need to make an additional assumption on the $$V_i$$'s to avoid this issue, but perhaps I'm mistaken.

For non-zero $$e$$, if $$\pi(e) \in U_i - V_i$$, then $$h_i'(e)$$ may or may not be zero depending on whether $$\lambda_i(\pi(e))$$ is zero or not. This is why it is useful to consider $$\{W_i\}$$ as, by construction, we have $$\lambda_i|_{W_i} \equiv 1$$.
Recall that we have covers $$\{W_i\}$$, $$\{V_i\}$$, and $$\{U_i\}$$ of $$B$$ such that $$W_i \subset V_i \subset U_i$$ with $$\overline{W_i} \subset V_i$$ and $$\overline{V_i} \subset U_i$$. For any $$e \in E(\xi)$$, we have $$\pi(e) \in W_i$$ for some $$i$$ since $$\{W_i\}$$ is a cover of $$B$$. As $$W_i \subset U_i$$ and $$\lambda_i|_{W_i} \equiv 1$$, we see that
$$h_i'(e) = \lambda_i(\pi(e))h_i(e) = h_i(e)$$
which is zero if and only if $$e$$ is the zero vector in the fiber over $$\pi(e)$$.
• Ah the $W_i$ are stipulated to form a cover of the base. This clears it up, thanks! Jan 10 at 21:13