# Pull-back of differential

Let $$f : U \rightarrow V$$ be a smooth map between open sets $$U \subset \mathbb{R}^{m}$$ and $$V \subset \mathbb{R}^{n}$$. Let us write $$f=\left(f_{1}, \ldots, f_{n}\right)$$ and let's use different variables for domain and codomain $$\left(x^{1}, \ldots, x^{n}\right) \in \mathbb{R}^{n}$$ e $$\left(y^{1}, \ldots, y^{m}\right) \in \mathbb{R}^{m}$$. I am asked to prove that $$f^{*}\left(d x^{i}\right)=\frac{\partial f_{i}}{\partial y^{j}} d y^{j}=d f_{i}$$ Let $$v \in T_p \mathbb{R}^{n} = \mathbb{R}^{n}$$. My proof goes like this $$\left(f^{*}\left(d x^{i}\right)\right)(p) (v) = d x^{i}_{f(p)} (df_p(v)) = d (x^{i}\circ f)_p(v) = d (f_i)_p (v)$$ and hence $$f^{*}\left(d x^{i}\right) = d f_{i} = \frac{\partial f_{i}}{\partial y^{j}} d y^{j}$$

However, I am supposed to prove $$f^{*}\left(d x^{i}\right)=\frac{\partial f_{i}}{\partial y^{j}} d y^{j}$$ directly. I managed to get $$\forall p \in U \ \forall v \in T_p U \ \forall g \in C^\infty (\mathbb{R})$$

\begin{align} \left(f^{*}\left(d x^{i}\right)\right)(p) (v) (g) & = d x^{i}_{f(p)} \left(df_p \left(\sum_j v^j \frac{\partial}{\partial y^j} \right) \right) (g) \\ & = \sum_j v^j \left( d (f_i)_p \left( \frac{\partial}{\partial y^j} \right) \right) (g) \\ & = \sum_j v^j \frac{\partial g \circ f_i}{\partial y^j} (p) \\ & = \sum_j dy^j_p(v) g'(f_i(p)) \frac{\partial f_i}{\partial y^j} (p) \end{align}

but I'm not really sure whether this is right. Does anyone know how to go on from here?

• What you're writing doesn't make sense. Evaluating a $1$-form on a tangent vector gives a number. It only makes sense to apply a tangent vector to a function ($g$) to compute a directional derivative. What does "directly" mean? If you know that $d$ commutes with pullback, that gives it to you directly: $f^*(dx^i) = d(f^*x_i) = df_i$. May 11, 2019 at 17:46
• Thanks for your comment! I was making use of a smooth function from $\mathbb{R}$ to $\mathbb{R}$ in order to apply the definition of differential $df_p(v)(g) = v(g\circ f)$. I am aware that a 1-form on a tangent vector gives a number, but I was hoping I could regard that number as an element of the tangent space of $\mathbb{R}$ through the canonic identification $a \mapsto a \frac{\partial}{\partial x}$ and use the definition of differential. Do you have some tips on how to compute $d (f_i)_p \left( \frac{\partial}{\partial y^j} \right)$ otherwise? May 11, 2019 at 18:20
• Nope, it's a number, not a tangent vector to $\Bbb R$. “Directly” should mean working just with differential forms (as I much prefer), rather than evaluating on tangent vectors. I gave you the argument, or are we trying to prove $d$ commutes with pullback? May 11, 2019 at 18:24
• Actually, I am not quite sure of what "directly" means as it is in the text of the problem which I am trying to solve but I interpreted it as "in coordinates" being the rhs in coordinates. May 11, 2019 at 19:56
• But let me see if I get this right: the differential of a map goes from a tangent space to a tangent space, so it maps vectors to vectors. A differential form is at each point is an antisymmetric tensor so yes I agree it gives me a number. However, I always thought I could interpret $df$ for a map $f: M \to \mathbb{R}$ as both a differential and a differential form because of the canonic identification from $\mathbb{R}$ to its tangent space which I wrote in the last message. Is this not correct? May 11, 2019 at 19:56