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.


1 Answer 1


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 $\pi(e)$.

  • $\begingroup$ Ah the $W_i$ are stipulated to form a cover of the base. This clears it up, thanks! $\endgroup$
    – Mr. Brown
    Jan 10 at 21:13

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