Pullback of Riemannian Metric by Smooth Mapping is Riemannian Metric iff Mapping is Immersion

Theorem

Let $\struct {\tilde M, \tilde g}$ be a Riemannian manifold with or without boundary.

Let $M$ be a smooth manifold with or without boundary.

Let $F : M \to \tilde M$ be a smooth mapping.

Let $F^* \tilde g$ be the pullback of $\tilde g$ by $F$.

Let $g = F^* \tilde g$ be a smooth $2$-tensor field.

Then $g$ is a Riemannian metric if and only if $F$ is an immersion.

Proof

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Sources

• 2018: John M. Lee: Introduction to Riemannian Manifolds (2nd ed.) ... (previous) ... (next): $\S 2$. Riemannian Metrics. Definitions