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

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?

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 $$F^*A$$?

• Well I haven't already studied exterior derivatives. Where do you use the fact you mention? In the equality $F^\ast (dy^{j})= d(y^jF)$? Commented Aug 19, 2020 at 13:21
• @arnett yes. $F^*(dy^j) = d(F^*y^j) = d(y^j \circ F)$; the proof is just a direct calculation by using the definition of $dy^j$ and $F^*$. Commented Aug 19, 2020 at 13:41