> [!definition] > > $ > \begin{CD} > X @>{F}>> Y \\ > @V{F^*\omega}VV @VV{\omega}V \\ > T^*X @<dF^*<< T^*Y > \end{CD} > $ > > Let $X$ and $Y$ be [[Manifold|manifolds]], and $F: X \to Y$ be a [[Manifold Morphism|morphism]]. Let $p \in X$, then the [[Differential|differential]] induces a [[Dual Map|dual map]] > $ > dF_p^*: T^*_{F(p)}Y \to T^*_pX > $ > known as the pointwise **pullback** by $F$ at $p$, or the **cotangent map** of $F$. If $\omega \in \vf^*(Y)$ is a rough covector field, then > $ > (F^*\omega)_p = dF_p^*(\omega_{F(p)}) > $ > is the **pullback** of $\omega$ by $F$. If $X$ and $Y$ are modelled on Euclidean Spaces, then the pullback has coordinate representation > $ > F^*\omega = (\omega_j \circ F)d(y^j \circ F) = (\omega_j \circ F)dF^j > $ > *Proof*. Write $\omega = \omega_j dy^j$, then > $ > \begin{align*} > F^*\omega &= F^*(\omega_jdy^j) \\ > &= (w_j \circ F)F^*dy^j \\ > &= (w_j \circ F)d(y^j \circ F) \\ > &= (w_j \circ F)dF^j > \end{align*} > $ > [!theorem] > > Let $u \in C(Y)$, and $\omega: Y \to T^*Y$ be a rough covector field, then > $ > F^*(u\omega) = (u \circ F)F^*\omega > $ > If $u \in C^p(Y)$ with $p \ge 1$, then > $ > F^*du = d(u \circ F) > $ > > *Proof*. > $ > \begin{align*} > F^*(u\omega)(p) &= dF_p^*(u\omega(F(p))) \\ > &= u(F(p)) \cdot dF_p^*(\omega(F(p))) \\ > &= (u \circ F)(p) \cdot F^*\omega(p) \\ > F^*(u\omega) &= (u \circ F) \cdot F^*\omega > \end{align*} > $ > If $u \in C^\infty(Y)$, then > $ > \begin{align*} > F^*du(p) &= dF_p^*(du_{F(p)}) \\ > &= du_{F(p)} \circ dF_p \\ > &= d(u \circ F)_p > \end{align*} > $ > [!theorem] > > Let $X$ and $Y$ be manifolds modelled on $E_1, E_2$ respectively. Let $\omega$ be a (continuous) covector field on $Y$. Then $F^*\omega$ is a continuous covector field on $X$. If $\omega$ is smooth, then so is $F^*\omega$. > > *Proof*. Let $(U, \varphi) \in X$ and $(V, \psi) \in Y$ be [[Atlas|charts]] with $F(U) \subset V$, and $\tau^*: T^*V \to V \times E^*$ be the trivialising map corresponding to $(V, \psi)$. Let $\wh F = F_{U, V}$ and $\td\omega = \proj_2 \circ \tau^* \circ \omega \circ \psi^{-1}$ be the local representations, the pullback has local representation > $ > \td{F^*\omega}: E_1 \to E_1 \times E_1^* > $ > where > $ > \td{F^*\omega} = \braks{D(\wh{F})(\hat p)}^* \cdot (\td{\omega} \circ \wh F(\hat p)) > $ > Since the composition between a vector and a map is bilinear and bounded, $\td{F^*\omega}$ is continuous if $\omega$ is continuous, and smooth if $\omega$ is smooth. > [!theorem] > > Let $(r, \theta)$ be the polar coordinates on the right half-plane $H = \bracs{(x, y): x > 0}$. Then the change of coordinates > $ > (x, y) = (r\cos \theta, r\sin \theta) > $ > is the identity map on $H$. Using $(r, \theta)$ as coordinates for the domain and $(x, y)$ as the codomain allows pulling back the differentials $dx$ and $dy$ as > $ > \begin{align*} > xdy - ydx &= I^*(xdy - ydx) \\ > &= r\cos\theta d(r\sin\theta) - r\sin\theta d(r\cos\theta) \\ > &= r\cos\theta\braks{\sin \theta dr + r\cos\theta d\theta} - r\sin\theta d\braks{\cos\theta dr - r\sin\theta d\theta} \\ > &= r\cos\theta \cdot \sin\theta - r\sin\theta \cos\theta dr + r^2\cos^2\theta d\theta + r^2\sin^2\theta d\theta \\ > &= r^2 d\theta > \end{align*} > $