I'm trying to understand the proposition which says that if $p: \mathcal{V} \rightarrow B$ is a real vector bundle and $j : B \rightarrow \mathcal{V}$ is the inclusion from B to $ \mathcal{V}$ as the zero section. Then $P^*: H^\bullet(B) \rightarrow H^\bullet( \mathcal{V})$ and $j^*: H^\bullet( \mathcal{V}) \rightarrow H^\bullet(B)$ are inverses to each other.
To do so I need to understand the proof of this lemma
Lemma 1.45. Let $\mathcal{R}$. be the vertical Euler vector field on $\mathcal{V}$, and let $h_t(v)=tv$. With $H:A^\bullet(\mathcal{V})\longrightarrow{A^\bullet}^{-1}(\mathcal{V})$ defined by the formula $$H\beta=\int_0^1h_t^*(\iota(\mathcal{R})\beta)t^{-1}dt.$$ we have the homotopy formula $$\beta-p^*(j^*\beta)=(dH+Hd)\beta,\quad\text{for all}\ \beta\in\mathcal{A(V)}.$$ Proof. First, note that the integral converges in the definition of $H$ because $\mathcal{R}$ vanishes at $0$. If $\beta\in\mathcal{A(V)}$, let $\beta_t=h_t^*\beta$, and observe that $\beta_0=p^*j^*\beta$ and that $\beta_1=\beta$. Differentiating by $t$ and using integrating the Cartan homotopy formula $$\mathcal{L(R)}\beta=d(\iota(\mathcal{R})\beta)+\iota(\mathcal{R})(d\beta),$$ we obtain the lemma.
My question are :
How to show that $\beta_0 = p^*j^*\beta$ ?
Why this is true $\frac{d}{dt}h^*_t \beta = t^{-1} \mathcal{L}(\mathcal{R})h^*_t\beta$.
Any help would be greatly appreciated.