# About the definition of $\Phi^*:\Omega^k(B)\longrightarrow \Omega^k(A)$?

Let $\Phi:A\longrightarrow B$ be a vector bundle morphism covering $\Phi_0:M\longrightarrow N$. Then there is an induced map $\Phi^*:\Omega^k(B)\longrightarrow \Omega^k(A)$ given by $$(\Phi^* \varepsilon)_p(a_1, \ldots, a_k):=\varepsilon_{\Phi_0(p)}(\Phi(a_1), \ldots, \Phi(a_k)).$$ I have seen in some places that $\Phi^*$ is defined like this: $$\langle \Phi^*\varepsilon, \alpha_1\wedge \ldots \wedge \alpha_k\rangle:=\langle \varepsilon\circ \Phi_0, \Phi(\alpha_1)\wedge \ldots \wedge \Phi(\alpha_k)\rangle,$$ where $\alpha_j\in \Gamma(A)$ (IMPORTANT: I didn't mistyped $\Gamma(A)$). Can anyone explain me this?

I don't understand why $\Phi^* \varepsilon$ is acting on $\alpha_1\wedge \ldots \wedge \alpha_k$ for $(\alpha_1\wedge \ldots \wedge \alpha_k)(p)\in \Lambda^k A_p$ and $(\Phi^* \varepsilon)_p$ should act on $\Lambda^k A_p^*$.

Thanks

Obs. $\Omega^k(A):=\Gamma(\Lambda^k A^*)$, that is, the elements of $\Omega^k(A)$ are sections of the bundle $\Lambda^k A^*\longrightarrow M$.

Consider the linear algebra picture. If $V$ is a vector space, an element of $\Lambda^k(V^{*})$ acts on $\Lambda^k(V)$ by the formula
$$(\varphi^1 \wedge \dots \wedge \varphi^k)(v_1 \wedge \dots \wedge v_k) = \det(\varphi^i(v_j))_{i,j=1}^k$$
(extended in a multilinear way to a pairing $\Lambda^k(V^{*}) \times \Lambda^k(V) \rightarrow \mathbb{R}$). This action allows you to identify elements of $\Lambda^k(V^{*})$ with multilinear alternating maps $$\underbrace{V \times \dots \times V}_{k\text{ times}} \rightarrow \mathbb{R}.$$
Now, the vector $\Phi^{*}(\varepsilon)_p$ should be an element of $\Lambda^k(A_p^{*})$ and so it should act on $\Lambda^k(A_p)$.