I am reading Guillemin and Pollack's Differential Topology Page 163:
If $f: X \to Y$ is a smooth map and $\omega$ is a $p$-form on $Y$, define a $p$-form $f^*\omega$ on $X$ as follows: $$f^*\omega(x) = (df_x)^*\omega[f(x)].$$
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Why it uses a pullback to define the pullback?
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What $\omega[f(x)]$ really mean?
Learning from answers below and more readings on the book, here's a fun expository about how I am trying to build my exterior algebra from definitions given in the book.
We use the pull-back in linear space on the right hand side to define the pull-back on the manifold, on the left hand side. To understand the linear space on the right, first note:
The collection of all $p$-tensors is a vector space $\mathcal{J}^p(V^*)$. (Page 154)
Then we see
The alternating $p$-tensors form a vector subspace $\Lambda^p(V^*)$ of $\mathcal{J}^p(V^*)$. (Page 156)
One more key-ingredient:
Suppose $A: V \to W$ is a linear map. Then the transpose map $A^*: W^* \to V^*$ extends to the exterior algebras, $A^*: \Lambda^p(W^*) \to \Lambda^p(V^*)$ for all $p>0$. If $T \in \Lambda^p(W^*)$, just define $A^* T \in \Lambda^p(V^*)$ by
$$A^*T(v_1, \dots, v_p) = T(Av_1, \dots, Av_p)$$
for all vectors $v_1, \dots, v_p \in V$. (Page 159)
Now we get everything we need, back to the question, the meaning of left hand side $(df_x)^*w[f(x)].$ The last definition below, gives us the connection:
Definition. Let $X$ be a smooth manifold with or without boundary, A $p$-form on $X$ is a function $\omega$ that assigns to each point $x \in X$ an alternating $p$-tensor $\omega(x)$ on the tangent space of $X$ at $x$; $\omega(x) \in \Lambda^p[T_x(X)^*].$ (Page 162)
Now everything unraveled.
Given $df_x: T_x(X) \to T_y(Y)$, we see $df_x$ as the linear map $A$ in definition on Page 159. Hence $V$ is $T_x(X)$ and $W$ is $T_y(Y)$. Then we obtain the transpose map $(df_x)^*: \Lambda^p[T_y(Y)]^* \to \Lambda^p[T_x(X)]^*$ for all $p>0$.
Furthermore, from Page 162, we see know $\omega$ is an alternating $p$-tensor, which is an element of $\Lambda^p(V^*)$ as the definition on Page 156. Note $\Lambda^p(V^*)$ is just $\Lambda^p(T_x(X)^*)$ as stated in the precede paragraph. But here calls the special attention: $\omega$ is not $T$. Because $\omega$ is in $\Lambda^p [T_x(X)^*]$ as the definition on Page 162, but $T$ is $\Lambda^p(W^*)$, which corresponds to $\Lambda^p [T_y(Y)^*]$.
Fortunately, we have $A^*: \Lambda^p(W^*) \to \Lambda^p (V^*)$, that in our question, is $df_x^*: \Lambda^p[T_y(Y)^*] \to \Lambda^p [T_x(X)^*]$. So now it is clear that $\omega$ is $(df_x)^*T$.
If $T \in \Lambda^p(T_y(Y)^*)$, just define $(df_x)^* T \in \Lambda^p(T_x(X)^*)$ by
$$(df_x)^*T(v_1, \dots, v_p) = T(df_x (v_1), \dots, df_x (v_p))$$
for all vectors $v_1, \dots, v_p \in T_x(X)$.
Basic linear algebra tells us
$$(df_x)^*T(v_1, \dots, v_p) = T \circ df_x (v_1, \dots, v_p)$$
From definition on Page 159, we know $A$ takes vectors in $V$ to $W$, and therefore $T$ takes the result of $A$, that is $W$. Hence in our case, $df_x$ takes vectors in $T_x(X)$ to $T_y(Y)$. Hence $T$ takes $T_y(Y)$, which is exactly $df_x (v_1, \dots, v_p)$!
All above is in the effort of making sense of $(df_x)^*\omega[f(x)].$ Now I have shown $(df_x)^*T(v_1, \dots, v_p) = T \circ df_x (v_1, \dots, v_p)$ and $\omega$ is $(df_x)^*T$. But still, there is a missing piece out there connects $\omega[f(x)]$ with these.
It seems the answer of $f: X \to Y$ is a smooth map and $\omega$ is a $p$-form on $Y$, what is $\omega[f(x)]$? mend the missing piece: $\omega$ is defined to be
$$\omega_\alpha(y) = \alpha(y)dx^{i_1}\wedge\ldots\wedge dx^{i_p}.$$
Hence $$\omega[f(x)] = (\alpha\circ f)(x)dx^{i_1}\wedge\ldots\wedge dx^{i_p}.$$
Therefore,
$$f^*\omega(x) = (df_x)^*\omega[f(x)] = (df_x)^*[(\alpha\circ f)(x)dx^{i_1}\wedge\ldots\wedge dx^{i_p}].$$
Best Answer
They use a pullback to define a pullback because $df_x$ is a linear map and so the pullback $df_x^\ast$ is already defined; they're using it to generalize to a manifold setting. And $w[f(x)]$ must mean $w$ at the point $f(x)$.