Pull-back of a smooth (covariant) tensor field is smooth

Question

I want to prove the statement in the title. To be more precise and fix the notation, let $F: X \to Y$ be a differentiable map between smooth manifolds and $A$ a $k$-covariant tensor field over $Y$. The pullback $F^\ast A$ is a $k$-covariant tensor field over $X$ defined by $$(F^\ast A)_{p}(S_1, \ldots, S_k)=A_{F(p)}(T_p S_1, \ldots , T_pS_k)$$ for every $S_1, \ldots, S_k$ in $T_pX$.

My idea was to work in local coordinates, but I'm doing something wrong. Suppose we can write $A$ locally as $$A=A_{j_1, \ldots, j_k} \,dy^{j_1} \otimes \ldots \otimes dy^{j_k}$$ in a coordinate chart $(V, \psi=(y^j)_{j=1}^n)$ with smooth coefficients $A_{j_1, \ldots, j_k}$, then I would like to prove that the coefficients of $F^\ast A$ are also smooth. But how to identify them (which is a problem of its own interest to me)? And also, is this general local expression for $A$ on $Y$ correct?

Answer 1

Pull-back of tensor fields plays nicely with the tensor product, and pull-back also plays nicely with the exterior derivative (make sure you prove these facts first), and pull-back of real-valued functions is simply composition: $F^*g = g \circ F$. Also, yes, that is the general local expression for $A$ on $Y$.

Now, if we take a chart $(U,\phi = (x^1,\dots, x^m))$ on $X$, such that $F(U)\subset V$ (or atleast $F(U)\cap V \neq \emptyset$) then \begin{align} F^*A &= F^*(A_{j_1 \dots j_k}) \cdot F^*(dy^{j_1}) \otimes \cdots \otimes F^*(dy^{j_k}) \\ &= (A_{j_1\dots j_k} \circ F) \cdot d(y^{j_1}\circ F) \otimes \cdots \otimes d(y^{j_k}\circ F) \\ &= (A_{j_1\dots j_k} \circ F) \cdot \left[\dfrac{\partial (y^{j_1}\circ F)}{\partial x^{i_1}} dx^{i_1}\right] \otimes \cdots \otimes \left[\dfrac{\partial (y^{j_k}\circ F)}{\partial x^{i_k}} dx^{i_k}\right] \\ &= \left[(A_{j_1\dots j_k} \circ F) \cdot \dfrac{\partial (y^{j_1}\circ F)}{\partial x^{i_1}} \cdots \dfrac{\partial (y^{j_k}\circ F)}{\partial x^{i_k}} \right] dx^{i_1}\otimes \cdots \otimes dx^{i_k} \end{align} (throughout, the $j$'s range over $\{1,\dots, n\}$ and $i$'s range over $\{1,\dots, m\}$, and summation convention is used).

Using smoothness of $A$ and $F$, what can you now deduce about smoothness of $F^*A$?