Prove the non-orientability of a vector bundle using the first Stiefel-Whitney class $w_1 \neq 0$ Question:
How to compute the first Stiefel-Whitney class $w_1$ of the following vector bundle? Or Prove it is $w_1\neq 0$.

*

*The base space is a torus $T^2$

*The fiber is locally $\mathbb{R}$

*The line bundle is non-trivial: along the each loop (no matter toroidal or poloidal), the fiber bundle is a mobius strip.

What I know:

*

*The first Stiefel-Whitney class of a fiber bundle $E$ is $w_1(E)\in H^1(M;\mathbb{Z}_2)$, where $M$ is the base space.


*The computing of $H^1(T^2;\mathbb{Z}_2)$
$\delta^0(p^*)=0$
$\delta(a^*,b^*,c^*)=（u^*,v^*）[ \begin{matrix}1 & 1 & 1\\1 & 1 & 1 \end{matrix}]$
$im\delta^0=0$
$ker\delta^1=\mathbb{Z}_2[a^*+b^*]\oplus \mathbb{Z}_2[b^*+c^*]$
$H^1(T^2)=ker \delta^1/im\delta^0=\mathbb{Z_2}[a^*+b^*]\oplus\mathbb{Z_2}[b^*+c^*]$

 A: Let $L \to S^1\times S^1$ denote the line bundle you describe.
The restriction of $L$ to $S^1\times\{p\}$ is non-trivial, so $L$ is non-trivial and hence $w_1(L) \neq 0$ since real line bundles are classified up to isomorphism by their first Stiefel-Whitney class. Identifying $w_1(L) \in H^1(S^1\times S^1; \mathbb{Z}_2)$ requires some more work.
For each $p \in S^1$ we have inclusions $i_p : S^1 \to S^1\times S^1$ and $j_p : S^1 \to S^1\times S^1$ given by $i_p(q) = (q, p)$ and $j_p(q) = (p, q)$ respectively. We also have projections $\pi_i : S^1\times S^1 \to S^1$ for $i = 1, 2$ where $\pi_i(a_1, a_2) = a_i$. Note that $\pi_1\circ i_p = \operatorname{id}_{S^1}$ and $\pi_2\circ j_p = \operatorname{id}_{S^1}$. On the other hand, $\pi_2\circ i_p = c_p$ and $\pi_1\circ j_p = c_p$ where $c_p : S^1\to S^1$ is the constant map given by $c_p(q) = p$.
Let $\gamma \to S^1$ denote the non-trivial real line bundle on $S^1$. From your description of $L$, we have $i_p^*L \cong \gamma$ and $j_p^*L \cong \gamma$.
By the Künneth theorem, there is an isomorphism
\begin{align*}
H^1(S^1; \mathbb{Z}_2)\oplus H^1(S^1; \mathbb{Z}_2) &\to H^1(S^1\times S^1; \mathbb{Z}_2)\\
(\alpha, \beta) &\mapsto \pi_1^*\alpha + \pi_2^*\beta.
\end{align*}
In particular, $w_1(L) \in H^1(S^1\times S^1; \mathbb{Z}_2)$ is of the form $\pi_1^*\alpha + \pi_2^*\beta$ for some $\alpha$ and $\beta$. Now note that
\begin{align*}
w_1(L) &= \pi_1^*\alpha + \pi_2^*\beta\\
i_p^*w_1(L) &= i_p^*\pi_1^*\alpha + i_p^*\pi_2^*\beta\\
w_1(i_p^*L) &= (\pi_1\circ i_p)^*\alpha + (\pi_2\circ i_p)^*\beta\\
w_1(\gamma) &= \operatorname{id}_{S^1}^*\alpha + c_p^*\beta\\
w_1(\gamma) &= \alpha.
\end{align*}
Likewise
\begin{align*}
w_1(L) &= \pi_1^*\alpha + \pi_2^*\beta\\
j_p^*w_1(L) &= j_p^*\pi_1^*\alpha + j_p^*\pi_2^*\beta\\
w_1(j_p^*L) &= (\pi_1\circ j_p)^*\alpha + (\pi_2\circ j_p)^*\beta\\
w_1(\gamma) &= c_p^*\alpha + \operatorname{id}_{S^1}^*\beta\\
w_1(\gamma) &= \beta.
\end{align*}
Therefore
$$w_1(L) = \pi_1^*\alpha + \pi_2^*\beta = \pi_1^*w_1(\gamma) + \pi_2^*w_1(\gamma).$$
Now that we have identified $w_1(L)$ and real line bundles are classified up to isomorphism by their first Stiefel-Whitney class, we can find an expression for $L$. Since
$$w_1(L) = \pi_1^*w_1(\gamma) + \pi_2^*w_1(\gamma) = w_1(\pi_1^*\gamma) + w_1(\pi_2^*\gamma) = w_1(\pi_1^*\gamma\otimes\pi_2^*\gamma),$$
we see that $L \cong \pi_1^*\gamma\otimes\pi_2^*\gamma$. For a proof of the last equality above, see the Lemma in this answer.
