Is the smooth embedding in the hypothesis of the Nash-Kuiper theorem required to be isometric?

Question

Wikipedia states the hypothesis of the Nash-Kuiper theorem as follows:

Let $(M,g)$ be an $m$-dimensional Riemannian manifold and $f: M \to \mathbb{R}^n$ a short $\mathcal{C}^\infty$-embedding (or immersion) into Euclidean space $\mathbb{R}^n$, where $n ≥ m+1$.

We usually either talk about (a) "embeddings" between differentiable manifolds or (b) "isometric embeddings" between Riemannian manifolds. So if someone talks about just an "embedding" between Riemannian manifolds, is it implied that this embedding is isometric, or does the embedding only apply to the underlying differentiable manifold structure and not to the metric structure?

I assume that $f$ isn't required to be isometric, because if it is then the Nash-Kuiper theorem seems totally trivial, because you can just let $f_\epsilon$ be $f$ itself.

Answer 1

No, the main point about the theorem is to prove the existences of an isometric embedding in the first place (which - as explained on the Wiki page, is a consequence of the theorem in question together with the Whitney embedding theorem, which is ignoring the metric structure). So the assumption in the Nash $C^1$ theorem is that you have an embedding in the sense of the underlying differentiable manifold structure.