Basic properties of push-forward and pullback
by Hidenori
Proposition
- If $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$ and $g: \mathbb{R}^m \rightarrow \mathbb{R}^p$, show that \((g \circ f)_{\ast} = g_{\ast} \circ f_{\ast}\) and $(g \circ f)^{\ast} = f^{\ast} \circ g^{\ast}$.
- If $f, g: \mathbb{R}^n \rightarrow \mathbb{R}$, show that $d(f \cdot g) = f \cdot dg + g \cdot df$.
Solution
1
Let $p \in \mathbb{R}^n$ be given. Let $v_p \in \mathbb{R}^n_p$ be given.
\[\begin{align*} (g \circ f)_{\ast}(v_p) &= (D(g \circ f)(p)(v))_{(g \circ f)(p)} \\ &= \Big(\big((Dg)(f(p)) \circ (Df(p))\big)(v)\Big)_{g(f(p))} \\ &= \Big((Dg)(f(p))((Df(p))(v))\Big)_{g(f(p))} \\ &= g_{\ast}\big(((Df(p))(v))_{f(p)}\big) \\ &= g_{\ast}(f_{\ast}(v_p)) \\ &= (g_{\ast} \circ f_{\ast})(v_p)). \end{align*}\]Let $k \in \mathbb{N}$ be given. Let $\omega \in \mathcal{T}^k(\mathbb{R}^p)$ be given.
\[\begin{align*} (g \circ f)^{\ast}(\omega)(v_1, \cdots, v_k) &= \omega((g \circ f)(v_1), \cdots, (g \circ f)(v_k)) \\ &= \omega(g(f(v_1)), \cdots, g(f(v_k))) \\ &= (g^{\ast} \omega)(f(v_1), \cdots, f(v_k)) \\ &= (f^{\ast} (g^{\ast} \omega))(v_1, \cdots, v_k) \\ &= ((f^{\ast} \circ g^{\ast})(\omega))(v_1, \cdots, v_k). \end{align*}\]Therefore, $(g \circ f)^{\ast} = f^{\ast} \circ g^{\ast}$.
2
Since $f, g$ are 0-forms, $f \cdot g = f \wedge g$ as mentioned on P.89[Spivak]. $d(f \wedge g) = df \wedge g + f \wedge dg$ by Theorem 4-10(2)[Spivak]. $df \wedge g = g \wedge df$ [P.79, Spivak]. Then $d(fg) = d(f \wedge g) = g \wedge df + f \wedge dg = gdf + fdg$.
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