# Exterior Derivative in overlapping charts

Given a smooth manifold $M$, say of dimension $n$ and two charts $(U,x)$ & $(V,y)$, I want to prove that if $U \cap V \neq \emptyset$, then the exterior derivatives $d_x$ and $d_y$ coincide in $U \cap V$.

For this, there is a Lemma (8.36) in Jeffrey Lee's Manifolds and Differential Geometry that says that if I have two natural graded derivations of the same degree defined on the same set (such as $d_x$ and $d_y$ in $U \cap V$), and they coincide when applied to smooth functions and exact forms, then they are equal.

If $\alpha$ is an exact form, then $\alpha = d\beta$ for some differential form $\beta$. The exterior derivative $d$ satisfies that $d \circ d = 0$, so $d_x\alpha = d_y\alpha = d \circ d \beta = 0$.

Now, given a smooth function $f \in C^\infty(U \cap V)$, I want to show that $d_xf = \sum_i \partial f/\partial x^i dx^i = \sum_i \partial f/\partial y^i dy^i = d_yf$, but I don't know how.

I only know that $\partial f/\partial x^i = \sum_k \partial f/\partial y^k \cdot \partial y^k/\partial x^i$.

\begin{align}d_yf &\\ &=\sum_i \frac{\partial f}{\partial y^i}dy^i \\ &=\sum_i\left(\sum_j \frac{\partial f}{\partial x^j}\frac{\partial x^j}{\partial y^i} \right)\left(\sum_k\frac{\partial y^i}{\partial x^k}dx^k\right) \\ &= \sum_{i}\left(\sum_{j,k} \frac{\partial f}{\partial x^j}\frac{\partial x^j}{\partial y^i}\frac{\partial y^i}{\partial x^k}dx^k\right) \\ &= \sum_{j,k} \frac{\partial f}{\partial x^j}\left(\sum_{i}\frac{\partial x^j}{\partial y^i}\frac{\partial y^i}{\partial x^k}\right)dx^k \\ &= \sum_{j,k} \frac{\partial f}{\partial x^j}\delta_{jk}dx^k\\ &=\sum_j \frac{\partial f}{\partial x^j}dx^j \\ &=d_xf.\end{align}
• Beautiful!, I missed that $dy^i = \sum_k \partial y^i/ \partial x^k dx^k$. Thanks – Luis Vera Oct 7 '15 at 11:52