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Definition of pull-back

Let $M$ and $N$ be two manifolds. The pull back of $f:M\to N$ is defined as

$$[(f^*\omega)(x)](\omega_1,\ldots,\omega_r)=\omega(f(x))\cdot(f'(x)\cdot\omega_1,\ldots,f'(x)\omega_r)$$

Where $\omega$ is a differential form of degree $r$ in $N$, $x\in M$ and $\omega_1,\ldots,\omega_r\in T_xM$.

Thus it takes a form $\omega\in N$ to a form $f^*\omega\in M$.

What I didn't understand

Let's take now a differential form $\omega$ of degree $r$ and class $C^1$ in $M$, i.e., $\omega$ is a function $x\in M\mapsto\omega(x)\in A_r(T_xM)$. The book I'm reading says for every parametrization $\varphi:U_0\to U$ in $M$, there is a unique form $d_{\varphi}\omega$ of degree $r+1$ in $U$ such that $\varphi^*(d_{\varphi}\omega)=d(\varphi^*\omega)$. (afterwards he defines this as being the exterior derivative $d\omega$).

So what is $\varphi^*$? the author of the book didn't define the pull-back of a parametrization. Is there some abuse of notation here?

I've been passing hours to try to understand this, I really need help to get this definition.

Remark: if you don't know/understand my notation, please leave a comment.

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    Why can't you apply the given definition to the map $\phi$? On manifolds, it's enough to work with everything locally... – 2017-02-25
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    @MichaelBurr Do you mean define a pull back which takes the forms of $U$ to the forms of $U_0$? – 2017-02-25
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    A parameterization appears to be a map, and you have the definition of a pull back for a map, so it seems like you should be able to combine the two definitions. You mention that the author didn't define the pull-back of a parametrization why can't you just think of a parametrization as a function? Yes, I mean the pull back which takes forms on $U$ to forms on $U_0$. – 2017-02-25
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    @MichaelBurr Ok, but there is another problem. Using this $\varphi^*$ as the pull-back you have just said, $d_{\varphi}\omega$ would be a differential form in the open set $U$, but he defines the exterior derivative of $\omega$ of the _manifold_ $M$ as $(d\omega)$ putting $(d\omega)(x)=(d_{\varphi}\omega)(x)$ for every $x\in M$. So $d_{\varphi}\omega$ must be a differential form in $M$ not in $U$. Did you get where I am stuck? Thank you – 2017-02-25
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    Differential forms are local structures. It is perfectly fine to define them on open subsets of a manifold. The trick is to then glue them together, in other words, you have to check that they agree whenever different open sets in an open cover overlap. – 2017-02-25
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    @MichaelBurr Ok, but how can we do this? See: a differential form $\omega$ of degree $r$ in an _open set_ $U\in \mathbb R^n$ is defined as a function $\omega:x\in U\mapsto \omega(x)\in A_r(\mathbb R^n)$ while the differential form $\omega$ of degree $r$ in a _manifold_ is defined as a function $\omega:x\in M\mapsto \omega(x)\in A_r(T_xM)$. I don't how these definitions overlap. **Notation**: $T_xM$ is the tangent vector space at the point $x\in M$ and $A_r(E)$ is the vector space of the alternating $r$ forms of the vector space $E$. – 2017-02-25
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    I think that you're confused about more than can be helped through a single question. I suggest that you get a big piece of paper and draw out where everything lives, what equality means in each space, etc. – 2017-02-25
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    I've already tried it and I didn't succeed. I will ask another question asking about the overlapping between these two definitions. Thank you! – 2017-02-25

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