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Let $(M,g)$ be a Riemannian manifold. By the Nash imbedding Theorem we can find an isometric embedding $\varphi:M\rightarrow\mathbb R^s$ for sufficiently large $s$.
In this sense we may view $M$ as a subset of $\mathbb R^s$.
Now here is my question:

For points $v\in\mathbb R^s$ which are close enough to $M$, how can we define the orthogonal projection of $v$ onto $M$?

If $M$ was a linear subspace of $\mathbb R^s$, then this would be clear.

In general this is not well defined, since there is no obvious way to project the origin onto the sphere.
But for points $v\in\mathbb R^s$ such that the distance $\mathrm{dist}(v,M)$ is very small, there should be a way to uniquely define this projection.

The sufficiently close condition should be understood as a condition such that the projection is well defined so that things like above cannot happen.
I think in general it is necessary to assume that $M$ is a closed subset of $\mathbb R^s$, but is there a way to define this projection without this assumption?

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The adequate concepts to make you idea precise are those of the normal bundle and of tubular neigborhoods of a smooth submanifold $M \subset \mathbb R^s$. See for example

John M. Lee, Introduction to Smooth Manifolds (p. 139 ff)

and

https://mathoverflow.net/q/283467.

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I think you do need $M$ to be closed for this to work. Consider $M$ the open unit disk (in $\mathbb{R}^2$) embedded into $\mathbb{R}^3$ like an almost closed taco, ie the points $(1,0)$ and $(-1,0)$ are mapped to the same point in $\mathbb{R}^3$ (note that they are not part of $M$). You don't get a well defined orthogonal projection around that point.

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  • $\begingroup$ but that embedding will not be isometric, right? The pullback of the standard metric on $\mathbb R^3$ will not agree with the metric on the unit disc. $\endgroup$ Aug 29, 2019 at 12:42
  • $\begingroup$ You can do this with a flat piece of paper, so the embedded disk is isometric to the disk in $\mathbb{R}^2$. $\endgroup$
    – quarague
    Aug 29, 2019 at 12:44
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    $\begingroup$ But even in this example, there’s a neighborhood of $M$ on which the projection is well defined. The neighborhood just won’t contain the problematic point. $\endgroup$
    – Jack Lee
    Aug 29, 2019 at 14:51
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    $\begingroup$ @JackLee You can't find a universal $\varepsilon$ so that all points with distance less than $\varepsilon$ are in the tubular neighborhood. The distance bound has to shrink to zero when you approach the problematic point. $\endgroup$
    – quarague
    Aug 30, 2019 at 6:35
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    $\begingroup$ @quarague — That’s true, but the OP didn’t ask for a neighborhood of uniform size. $\endgroup$
    – Jack Lee
    Aug 30, 2019 at 13:15

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