# Thursday, October 29

:::{.exercise title="?"}
Prove that $d(f^* \alpha) = f^*(d\alpha)$ from the definition of $d$.
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:::{.exercise title="?"}
In local coordinates $x_1, \cdots, x_m$ on $M$ and $y_1, \cdots, y_n$ on $N$, where locally $f:M\to N$ is given by $y_1 = f_1(x_1, \cdots, x_n), y_2 = f_2(x_1, \cdots, x_n), \cdots$, set $\alpha =  \sum_{i_1 < \cdots < i_k} a_{i_1, \cdots, i_k} \bigvee_j dx_{i_j}$ and find a formula for $f^* \alpha = \sum b_I dx_I$.
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:::{.exercise title="?"}
Recall that if $\omega\in \Omega^k(M)$, we have
\[  
d\omega(v_0, \cdots, v_k) = \sum (-1)^i D_{v_i}\qty{\omega(v_0, \cdots, \hat v_i, \cdots, v_k) }
+ \sum_{i< j} (-1)^{i+j} \omega\qty{ [v_i, v_j], v_0, \cdots, \hat v_i, \cdots, \hat v_j, \cdots, v_k }
.\]

Show that this is independent of the choice of extension $v_0, \cdots, v_k \in T_p M$ to vector fields on $M$.

> Hint: consider the difference of two such choices and evaluate at $p$.


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