defining a pullback of differential forms

Does this definition of a pullback of a differential form make sense?

$\phi:M\rightarrow N$ and $\alpha\in\Omega^r(N)$ then define $$(\phi^*\alpha(X_1,\dots,X_r))(p) := \alpha(\phi(p))(\phi_{*_p}X_1(p),\dots,\phi_{*_p}X_r(p))$$ If it does make sense, it is annoying because it mixes both interpretations of a differential form. Ie on one side it is a map from vector fields to smooth functions and on the other, it is a map from a manifold to tangent bundle.

• In the last sentence, I think you mean a map from a manifold to some exterior power of the cotangent bundle. – Michael Albanese Dec 6 '17 at 16:36

$$(\phi^*\alpha)(p)(X_1(p), \dots, X_r(p)) = \alpha(\phi(p))(\phi_{*_p}X_1(p), \dots, \phi_{*_p}X_r(p)).$$
• for the first line, $\phi_*X$ is not necessarily a vector field on the whole of the manifold. Is this just solved by saying that $\alpha$ is local? – tomak Dec 7 '17 at 10:23
• @tomak: You're right, I was being silly. The expression $\phi_*X$ need not be well-defined (for example, when $\phi(p) = \phi(q)$ but $\phi_{*_p}X(p) \neq \phi_{*_q}X(q)$). I will remove it. – Michael Albanese Dec 7 '17 at 15:05