In Bott & Tu's "Differential forms", Example 10.1 states:

$\textbf{Example 10.1}$ Let $\pi: E \to M$ be a fiber bundle with fiber $F$. Define a presheaf on $M$ by $\mathcal F(U) = H^q(\pi^{-1}(U))$. For $U$ contractible, $H^q(\pi^{-1} U) \cong H^{q}(F)$ by the Kunneth formula, and if $V \subseteq U$ with $V$ connected, then the restriction $\rho^U_V: H^q(\pi^{-1}(U) \to H^q(\pi^{-1} V)$ is the identity. Therefore $\mathcal F$ is a locally constant presheaf on $M$.

However, consider the trivial rank $1$ bundle on $\mathbb R^2$. For a connected subset $V (\cong S^1 \times \mathbb R)\subseteq D^2$, we have $\mathcal F(V) = H^q(S^1) \ncong \mathcal F(U) = H^q(\text{point})$.

What am I missing here?

In Bott & Tu's "Differential forms", Example 10.1 states:

$\textbf{Example 10.1}$ Let $\pi: E \to M$ be a fiber bundle with fiber $F$. Define a presheaf on $M$ by $\mathcal F(U) = H^q(\pi^{-1}(U))$. For $U$ contractible, $H^q(\pi^{-1} U) \cong H^{q}(F)$ by the Kunneth formula, and if $V \subseteq U$ with $V$ connected, then the restriction $\rho^U_V: H^q(\pi^{-1}(U)) \to H^q(\pi^{-1} (V))$ is the identity. Therefore $\mathcal F$ is a locally constant presheaf on $M$.

However, consider the trivial rank $1$ bundle on $\mathbb R^2$. For a connected subset $V \cong S^1 \times \mathbb R \subseteq D^2$, we have $\mathcal F(V) = H^q(S^1) \ncong \mathcal F(U) = H^q(\text{point})$.

What am I missing here?