# Pullback map and its equality consequence

On my textbook, it says:

$$F^{\star}(dy^i) = \sum_{j=1}^{n} \frac{\partial y_i}{dx_j}dx_j$$

where $F^{\star}$ is a pullback map, map $F: M_1 \rightarrow M_2$ and $dx_j$s are forms on $M_1$ and $dy_j$ are forms on $M_2$.

The question is, isn't $dy_i$ itself already equals to $\sum_{j=1}^{n} \frac{\partial y_i}{dx_j}dx_j$? So what is $F^{\star}$ exactly doing then?

## 2 Answers

No; $dy^i$ is defined in (the cotangent bundle of ) $M_2$ , so that it cannot be written in terms of the basis {$dx^j$} of $M_1$ , if $M_1 \neq M_2$. The pullback allows you to transport the form from being defined in $TM_2$ ( or, eqiovalently, in the tangent spaces at points in $M_2$ ) into being defined in $TM_1$. This is an application of the general case that an n-linear map $B:V_1\times... V_n \rightarrow W_1\times... W_n$ between finite-dimensional vector spaces gives rise to (the dreaded, infamously-overused "ïnduces") an n-linear map $B^*$

$B^*:( W_1\times... W_n)^* \rightarrow (V_1\times... V_n)^*$

in a "functorial"way, i.e., the assignment of the ïnduced" multilinear map $B^*$ to the original bilinear map B is a functor between categories of vector spaces.

So, $(x_j)_j:U_1\to\Bbb R^n$ is a local chart on $U_1\subseteq M_1$ and $(y_i)_i$ on $U_2\subseteq M_2$. The point is that $\displaystyle\frac{\partial y_i}{\partial x_j}$ is (most probably) meant as $$\frac{\partial ( y_i\circ F)}{\partial x_j}\,.$$ (So, at a point $p_1\in M_1$, its '$y_i$ value' can be interpreted only via $F$, as $y_i(F(p_1))$.)