Let $(u, v)\in\mathbb{R}^2$ and consider the 1-form $\alpha_{(u,v)} : \mathrm{T}_{(u, v)}\mathbb{R}^2 \to \mathbb{R}$ defined by \begin{align} \alpha_{(u,v)}(w) &= (e^v ~\mathrm{d}u + u ~\mathrm{d}v) \cdot w \\ &= e^v \cdot w_1 + u \cdot w_2 \\ w &= w_1 \frac{\partial }{\partial u} + w_2 \frac{\partial}{\partial v} \end{align} Consider additionally the function $\phi : \mathbb{R}^3\to\mathbb{R}^2$ defined by $\phi(x, y, z) = (x+z, xy)$. I wish to compute the pullback of $\alpha$ by $\phi$.
Here is my attempt. I am wondering if I have done this correctly. The pullback is defined by, \begin{align} \phi^*\alpha(r) = \alpha_{\phi(x, y,z)}(\mathrm{T}_{(x, y,z)} \phi \cdot r) \end{align} where $\mathrm{T}_{(x, y,z)} \phi: \mathrm{T}_{(x, y,z)}\mathbb{R}^3 \to \mathrm{T}_{\phi(x, y,z)}\mathbb{R}^2$ is the derivative map and $r\in \mathrm{T}_{(x,y,z)}\mathbb{R}^3$. I compute the Jacobian of $\phi$ as \begin{align} \mathrm{T}_{(x, y,z)}\phi = \begin{pmatrix} 1 & 0 & 1 \\ y & x & 0 \end{pmatrix} \end{align} Writing $r = r_1 \frac{\partial}{\partial x} + r_2 \frac{\partial }{\partial y} + r_3 \frac{\partial}{\partial z}$ I compute, \begin{align} \phi^*\alpha(r) &= \alpha_{(x+z, xy)}((r_1 + r_3)~\mathrm{d}u + (yr_1 + x r_2) ~\mathrm{d}v) \\ &= e^{xy} (r_1 + r_3) + (x+z) (y r_1 + xr_2). \end{align} Have I done this correctly? Anything that could be improved?
A commenter suggested to write this in the form $\phi^*\alpha = a ~\mathrm{d}x + b ~\mathrm{d}y + c~\mathrm{d}z$. By identification, we have, \begin{align} \phi^*\alpha = (e^{xy} + y(x+z)) ~\mathrm{d}x +x (x+z) ~\mathrm{d}y + e^{xy} ~\mathrm{d}z. \end{align}