# Pullback of $dx_i$ in $\mathbb{R}^l$

I am trying to understand a line of reasoning in Guillemin and Pollack. Let $U\subset \mathbb{R}^k$ and $V\subset \mathbb{R}^l$ be open sets, and let $f:V\rightarrow U$ be smooth. Use $x_1,...,x_k$ for the standard coordinate functions on $\mathbb{R}^k$ and $y_1,...,y_l$ on $\mathbb{R}^l$.

I am trying to understand the following equation: $f^*dx_i=\sum_{j=1}^l\frac{\partial f_i}{\partial y_j}dy_j=df_i$, where $f^*$ denotes the pullback by $f$.

For a vector $v\in V$, we have $f^*dx_i(v)=dx_i(df(v))$, but I'm not sure where to go from here. I know that $df$ pushes the vector forward into $\mathbb{R^k}$, and that $dx_i$ in turn measures the $i$th coordinate of that vector; however, I am confused why it should be equal to the some on the right. How are they equal?

• Are you sure that $f^*dx_i(v)=dx_i(df(v))$? I would say the right equation is $f^*dx_i(v)=df^*(x_i)(v)$ which is equal to $d(x_i\circ f)(v)=df_i(v)=\sum_{j=1}^l\frac{\partial f_i}{\partial y_j}dy_j(v)$ – Javi Apr 18 '18 at 21:53
• The pullback in general is defined as: if $f:X\rightarrow Y$ is a smooth map, $f(x)=y$, and $w$ is a p-form on $Y$, then $f^*w=(df_x)^*w[f(x)]$ – ponchan Apr 18 '18 at 21:56
• Because the $\frac{\partial}{\partial y_i}$ form a basis of the tangent space and $Y$ is a tangent vector. – Javi Apr 18 '18 at 22:01
• It's not a power, it's just a common notation in this field to write the components, you can write them as subindices if you dare – Javi Apr 18 '18 at 22:05
• Strictly speaking you're right, it is implicit that $Y$ and $\frac{\partial}{\partial y_i}$ are being evaluated at some point. – Javi Apr 18 '18 at 22:09