If the first Stiefel-Whitney class of a subset $\gamma$ of a vector bundle $E$ is $w_1(\gamma)\neq 0$, is $w_1(E)\neq 0$ as well?

Question

Define:

• A vector bundle $E$ over a m-dimension base space $M$, $f: E\rightarrow M $
• A non-contractible loop $L$ in the base space $M$
• The vector bundle of the loop $L$ is $\gamma$ , $g: \gamma \rightarrow L$

Question:

• If the first Stiefel-Whitney class of the vector bundle bundle $\gamma$ is $w_1(\gamma)\neq0$, do we have $w_1(E)\neq0$?

Answer 1

The restriction of a vector bundle $\pi : E \to M$ to $L \subseteq M$ is $E|_L := \pi^{-1}(L)$.

If $i : L \to M$ denotes the inclusion map, then $E|_L \cong i^*E$. To see this, note that the total space of the pullback bundle is given by $i^*E = \{(\ell, e) \in L\times E \mid i(\ell) = \pi(e)\}$, see here. An isomorphism $i^*E \to E|_L$ is given by $(\ell, e) \mapsto e$ with inverse given by $e \mapsto (\pi(e), e)$. With this in mind, we have

$$w_1(E|_L) = w_1(i^*E) = i^*w_1(E).$$

So if $w_1(E) = 0$, then $w_1(E|_L) = i^*w_1(E) = i^*0 = 0$. Therefore, if $w_1(E|_L) \neq 0$, then $w_1(E) \neq 0$.