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False statementConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-10
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The pullback of a riemannian metric by every smooth map is a riemannian metric

Statement

Every smooth map pulls a Riemannian metric back to a Riemannian metric.

Facts & Assumptions

Given: The proposed universal claim; take F:RR, F(x)=0, with target metric h=dy2.

[F1]

Pullback of a riemannian metric as a tensor: For smooth F:MN and a Riemannian metric h on N, its pullback tensor is (Fh)p(v,w)=hF(p)(dFpv,dFpw). This is def-pullback-of-a-covariant-tensor-field for the tensor in def-riemannian-metric-and-riemannian-manifold. It is always symmetric and positive semidefinite; the name does not assert positive definiteness. Smoothness and the precise immersion criterion are established next.

[F2]

Pullback of a riemannian metric is riemannian exactly for immersions: Fh is Riemannian if and only if F is an immersion. In general it is positive semidefinite, with radical kerdFp at p.

Refutation

technique · direct
1.1

The coordinate function of F is constant, hence smooth with dFx(v)=0 for every x,vR. The target quadratic form is hy(w,w)=w2>0 for w0, so the target is Riemannian.

given
2.1

The pullback definition gives (Fh)x(x,x)=h0(0,0)=0. Since x0, positive definiteness fails; equivalently this F is not an immersion. Thus this smooth map refutes the claim.

F1F2step 1.1

Source locator

Lee, Introduction to Smooth Manifolds, 2nd ed., pp. 330–331, pullback metrics and Proposition 13.9; the constant-map computation above is explicit.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources