PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01

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Pullback of covariant tensors is smooth and functorial

Statement

If $F : M \to N$ is smooth and $T$ is a smooth covariant tensor field on $N$ , then $F^{*} T$ is a smooth covariant tensor field on $M$ . Moreover,

$({id}_{M})^{*} T = T, (G \circ F)^{*} T = F^{*} (G^{*} T)$

for every composable smooth map $G$ .

Facts & Assumptions

Given: Smooth maps $F : M \to N$ and $G : N \to P$ , and a smooth covariant tensor field $T$ on the target.

[F1]

Pullback of a covariant tensor field is defined by inserting the differential into every slot (The pullback of a covariant tensor field).

[L1]

Tensor pullback commutes with the corresponding fibrewise linear operations, and tensor products and contractions of smooth tensor fields are smooth (Linear pullback respects tensor products and permutations, Tensor products and contractions of smooth tensor fields are smooth).

[L2]

Differentials satisfy the chain rule (The chain rule for differentials of smooth maps).

Proof

technique · direct

1.1F1L1givenalgebra

In local coordinates, [F1] expresses each coefficient of $F^{*} T$ as a finite sum of the coefficients of $T \circ F$ multiplied by partial derivatives of $F$ . Those are smooth, so [L1] implies that $F^{*} T$ is smooth.

1.2F1given

The identity map has identity differential, so [F1] gives $({id}_{M})^{*} T = T$ .

1.3F1L2givenalgebra

For $v_{1}, \dots, v_{k} \in T_{p} M$ , [F1] and [L2] give $((G \circ F)^{*} T)_{p} (v_{1}, \dots, v_{k}) = T_{G (F (p))} (d G_{F (p)} d F_{p} v_{1}, \dots, d G_{F (p)} d F_{p} v_{k}),$ which is exactly $(F^{*} (G^{*} T))_{p} (v_{1}, \dots, v_{k})$ .

2.1step 1.1step 1.2step 1.3∎

Therefore pullback of covariant tensors is smooth and functorial.

Depends on

Dependency tree · 2 levels

Sources

John M. Lee, Introduction to Smooth Manifolds, 2nd ed.