False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-09-01

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A general mixed tensor field does not have a pullback by every smooth map

Statement

False claim: for every smooth map $F : M \to N$ and every vector field $Y$ on $N$ , there is a vector field $X$ on $M$ satisfying

$d F_{p} (X_{p}) = Y_{F (p)} (p \in M) .$

Such an $X$ would be the natural candidate for a pullback of $Y$ along $F$ .

Facts & Assumptions

Given: The smooth map $F : R \to R^{2}$ , $F (t) = (0, 0)$ , and the constant vector field $Y = \partial / \partial x$ on $R^{2}$ .

[L1]

Covariant tensor fields do admit functorial pullbacks (Pullback of covariant tensors is smooth and functorial).

Refutation

technique · direct

1.1

The field $Y$ is a type $(1, 0)$ tensor field, so the false claim requires a vector field $X$ on $R$ with $d F_{t} (X_{t}) = Y_{F (t)}$ for every $t$ .

given

2.1

The differential of the constant map is zero at every point, so $d F_{t} (X_{t}) = 0$ for every possible $X_{t} \in T_{t} R$ , whereas $Y_{F (t)} \neq 0$ . Thus the required equality is impossible.

givenstep 1.1algebra

3.1

Thus a general mixed tensor field does not have a pullback by every smooth map, even though [L1] shows that purely covariant tensors do.

L1step 2.1∎

Depends on

Pullback of covariant tensors is smooth and functorial

Used by

A vector field with no pullback under a noninjective map Counterexample

Dependency tree · two levels

5 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Will J. Merry, Differential Geometry (standard reference, not scraped)