How to calculate the pullback of a $k$-form explicitly I'm having trouble doing actual computations of the pullback of a $k$-form. I know that a given differentiable map $\alpha: \mathbb{R}^{m} \rightarrow \mathbb{R}^{n}$ induces a map $\alpha^{*}: \Omega^{k}(\mathbb{R}^{n}) \rightarrow \Omega^{k}(\mathbb{R}^{m})$ between $k$-forms. I can recite how this map is defined and understand why it is well defined, but when I'm given a particular $\alpha$ and a particular $\omega \in \Omega^{k}(\mathbb{R}^{n})$, I cannot compute $\alpha^{*}\omega$. 
For example I found an exercise (Analysis on Manifolds, by Munkres) where $\omega = xy \, dx + 2z \, dy - y \, dz\in \Omega^{k}(\mathbb{R}^{3})$ and $\alpha: \mathbb{R}^{2} \rightarrow \mathbb{R}^{3}$ is defined as $\alpha(u, v) = (uv, u^{2}, 3u + v)$, but I got lost wile expanding the definition of $\alpha^{*} \omega$. How can I calculate this?
Note: This exercise is not a homework, so feel free to illustrate the process with any $\alpha$ and $\omega$ you wish.
 A: Instead of thinking of $\alpha$ as a map, think of it as a substitution of variables:
$$
x = uv,\qquad y=u^2,\qquad z =3u+v.
$$
Then
$$
dx \;=\; \frac{\partial x}{\partial u}du+\frac{\partial x}{\partial v}dv \;=\; v\,du+u\,dv
$$
and similarly
$$
dy \;=\; 2u\,du\qquad\text{and}\qquad dz\;=\;3\,du+dv.
$$
Therefore,
$$
\begin{align*}
xy\,dx + 2z\,dy - y\,dz \;&=\; (uv)(u^2)(v\,du+u\,dv)+2(3u+v)(2u\,du)-(u^2)(3\,du+dv)\\[1ex]
&=\; (u^3v^2+9u^2+4uv)\,du\,+\,(u^4v-u^2)\,dv.
\end{align*}
$$
We conclude that
$$
\alpha^*(xy\,dx + 2z\,dy - y\,dz) \;=\; (u^3v^2+9u^2+4uv)\,du\,+\,(u^4v-u^2)\,dv.
$$
A: In some of my research I encountered $2$-forms that were given by $$\omega = dx \wedge dp + dy \wedge dq$$
and a map $$i : (u,v) \mapsto (u,v,f_u,-f_v)$$
for a general smooth map $f : (u,v) \mapsto f(u,v)$. I wanted to calculate the pullback of this map, i.e. $i^*\omega$. So,
\begin{align}
i^*\omega &= i^*(dx \wedge dp + dy \wedge dq) \\
&= d(x \circ i)\wedge d(p \circ i) + d(y \circ i)\wedge d(q \circ i).
\end{align}
Now, calculating each terms gives
\begin{align}
d(x \circ i) &= d(u) = du, \\
d(y \circ i) &= d(v) = dv, \\
d(p \circ i) &= d(f_u) = f_{uu}du + f_{uv}dv, \\
d(q \circ i) &= d(-f_v) = -f_{vu}du - f_{vv}dv.
\end{align}
Then, the pullback is given by
\begin{align}
i^*\omega &= du \wedge (f_{uu}du + f_{uv}dv) - dv \wedge (f_{vu}du + f_{vv}dv) \\
&= du \wedge (f_{uu}du) + du \wedge (f_{uv}dv) - dv \wedge (f_{vu}du) - dv \wedge (f_{vv}dv).
\end{align}
Now here, since the wedge product is $C^\infty(M)$-bilinear rather than just $\Bbb R$-bilinear, and $du \wedge dv = -dv\wedge du$. Using this property we have 
$$i^*\omega =2f_{uv} du \wedge dv.$$
I realise that these are based solely on $2$-forms and you were asking about a general $k$-form but you did write pick any $\alpha$ and $\omega$!
A: An answer that uses the definitions. I tried to write down everything with the hope to clarify the ideas. I do not know if this is helpful or simply too  verbose...
If $\omega\in\Omega^1(N)$ and $\alpha:M\rightarrow N$. The aim of the pullback is to define a form  $\alpha^*\omega\in\Omega^1(M)$ from a form $\omega\in\Omega^1(N)$.
A 1-form $\omega$ evaluated at $n=(x,y,z)\in N$ is $$\omega[n]=\omega_x(n)dx+\omega_y(n)dy+\omega_z(n)dz$$ where 


*

*$(\omega_x(n),\omega_y(n),\omega_z(n))\in\mathbb{R}^3$

*$dx, dy, dz$ are also 1-forms, the dual basis of $\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}$.


$\omega[n]\in T_nN^*$ eats one vector $b\in T_nN$ (a tangent vector at point $n$): 
\begin{align*}
\omega[n](b^x\frac{\partial}{\partial x}+b^y\frac{\partial}{\partial y}+b^z\frac{\partial}{\partial z}) &=\omega_x(n)b^x+\omega_y(n)b^y+\omega_z(n)b^z\in\mathbb{R}
\end{align*}
Now the objective is to define $\alpha^*\omega\in\Omega^1(M)$ from $\omega\in\Omega^1(N)$. Given  a point $m=(u,v)\in M$ and a vector $a\in T_m M$, a natural idea is to use the point $n=\alpha(m)\in N$ and to pushforward a vector $a\in T_mM$ to get a vector $\alpha_*(a)\in T_{\alpha(m)}N$. Actually this is how $\alpha^*\omega$ is defined:
$$(\alpha^*\omega)[m](a)=\omega[\alpha(m)](\alpha_*(a))$$
Now we must compute $\alpha_*(a)\in T_{\alpha(m)}N$, see at the end for details, we get:
$$
\alpha_*(a)=d\alpha^x[\alpha(m)](a)\frac{\partial}{\partial x}+d\alpha^y[\alpha(m)](a)\frac{\partial}{\partial y}+d\alpha^z[\alpha(m))](a)\frac{\partial}{\partial z}
$$
Thus:
\begin{align*}
(\alpha^*\omega)[m](a)&=\overbrace{\omega_x[\alpha(m)]d\alpha^x[\alpha(m)](a)}^{\text{term }dx(\frac{\partial}{\partial x})=1}+
\overbrace{\omega_y[\alpha(m)]d\alpha^y[\alpha(m)](a)}^{\text{term }dy(\frac{\partial}{\partial y})=1}+
\overbrace{\omega_z[\alpha(m)]d\alpha^z[\alpha(m)](a)}^{\text{term }dz(\frac{\partial}{\partial z})=1} \in \mathbb{R}
\end{align*}
We can drop the argument, vector $a=a^u\frac{\partial}{\partial u}+a^v\frac{\partial}{\partial v}$, it remains:
$$
(\alpha^*\omega)[m]=\omega_x[\alpha(m)]d\alpha^x[\alpha(m)]+
\omega_y[\alpha(m)]d\alpha^y[\alpha(m)]+
\omega_z[\alpha(m)]d\alpha^z[\alpha(m)]\in T_mM^*
$$
In the example: $(\omega_x,\omega_y,\omega_z)=(xy,2z,-y)$ and $\alpha: (u,v)\mapsto (uv,u^2,3u+v)$, therefore:


*

*$(\omega_x[\alpha(m)],\omega_y[\alpha(m)],\omega_z[\alpha(m)])=(u^3v,6u+2v,-u^2)$

*$d\alpha^x[\alpha(m)]=vdu+udv$

*$d\alpha^y[\alpha(m)]=2udu$ 

*$d\alpha^z[\alpha(m)]=3du+dv$
By substitution we get the expected result:
$$
(\alpha^*\omega)[m]=(u^3v)(vdu+udv)+(6u+2v)(2udu)+(-u^2)(3du+dv)
$$
$$
\alpha^*\omega=(u^3v^2+9u^2+4uv)du+(u^4v-u^2)dv\in\Omega^1(M)
$$

It remains to explain how to compute $\alpha_\star(a)\in T_{\alpha(m)}N$ from a tangent vector $a\in T_mM$
One can interpret a vector $a\in T_mM$ as a first order differential operator acting on functions $f:M\to\mathbb{R}$.
More explicitly we have:
$$
a[f]=(a^u\frac{\partial}{\partial u}+a^v\frac{\partial}{\partial v})f=df[m](a) \in\mathbb{R}
$$
The pushforward $\alpha_*$ transform a vector $a\in T_mM$ into a vector $\alpha_\star(a)\in T_{\alpha(m)}N$, thus it must act on functions $g:N\to\mathbb{R}$. A natural definition is:
$$
\alpha_\star(a)[g]=d(g\circ\alpha)[m](a)\in\mathbb{R}
$$ 
Expanding this formula we get:
\begin{align*}
d(g\circ\alpha)[m](a)&=\left((\frac{\partial g}{\partial x}\frac{\partial \alpha^x}{\partial u}+\frac{\partial g}{\partial y}\frac{\partial \alpha^y}{\partial u}+\frac{\partial g}{\partial z}\frac{\partial \alpha^z}{\partial u})du+(\frac{\partial g}{\partial x}\frac{\partial \alpha^x}{\partial v}+\frac{\partial g}{\partial y}\frac{\partial \alpha^y}{\partial v}+\frac{\partial g}{\partial z}\frac{\partial \alpha^z}{\partial v})dv\right)(a) \\
&= \left((\frac{\partial \alpha^x}{\partial u}du+\frac{\partial \alpha^x}{\partial v}dv)\frac{\partial g}{\partial x}+(\frac{\partial \alpha^y}{\partial u}du+\frac{\partial \alpha^y}{\partial v}dv)\frac{\partial g}{\partial y}+(\frac{\partial \alpha^z}{\partial u}du+\frac{\partial \alpha^z}{\partial v}dv)\frac{\partial g}{\partial z}\right)(a) \\
&= (\frac{\partial \alpha^x}{\partial u}du+\frac{\partial \alpha^x}{\partial v}dv)(a)\frac{\partial g}{\partial x}+(\frac{\partial \alpha^y}{\partial u}du+\frac{\partial \alpha^y}{\partial v}dv)(a)\frac{\partial g}{\partial y}+(\frac{\partial \alpha^z}{\partial u}du+\frac{\partial \alpha^z}{\partial v}dv)(a)\frac{\partial g}{\partial z} \\
&= \left( d\alpha^x[m](a)\frac{\partial}{\partial x}+d\alpha^y[m](a)\frac{\partial}{\partial y}+d\alpha^z[m](a)\frac{\partial}{\partial z} \right) g
\end{align*}
we get the expected result:
$$
\alpha_\star(a)=d\alpha^x[m](a)\frac{\partial}{\partial x}+d\alpha^y[m](a)\frac{\partial}{\partial y}+d\alpha^z[m](a)\frac{\partial}{\partial z}\in T_{\alpha(m)}N
$$
CAVEAT: we can always pullback differential forms, but only pushforward vectors (and not vector fields, unless $\alpha$ is a diffeomorphism (which is obviously not the case here)). See wikipedia, pushforward for further details.
