I am trying to understand a proof in Lee - Introduction to Smooth Manifolds. I have exactly the same question as was asked here: A submanifold is embedded iff it satisfies the local k-slice condition. The answer there says the containment $\iota(U_0)\subseteq V_0$ is clear, but I cannot see it. It is clearly true in coordinates, since for $u \in U_0$ we have that $\iota(u) = (u^1, \dots, u^k,0, \dots, 0) \in \iota(U_0)$ is also an element of $V_0$. However, I cannot see how to conclude the result.

I am in particular worried about the fact that all balls in Euclidean space are diffeomorphic to each other. Therefore the fact that $U_0$ and $V_0$ both have radius $\varepsilon$ in local coordinates does not seem sufficient to guarantee that $\iota(U_0) \subseteq V_0$.

  • $\begingroup$ I might be reading this wrong but I don't think you choose the same $\epsilon$ for both. You have some $\epsilon$ so that $U_0$ is an open ball. Then there is some open ball in $V$ that contains $i(U_0)$, which we call $V_0$. So $i(U_0) \subset V_0$ gives a slice chart. $\endgroup$ Commented Jul 23, 2020 at 0:49
  • $\begingroup$ @OsamaGhani I suppose I should have rewritten the question for myself instead of linking. In any case, Lee is clear that there is a single $\varepsilon$ chosen as the radius of both coordinate balls. $\endgroup$
    – Victor
    Commented Jul 23, 2020 at 0:58
  • $\begingroup$ I pulled up my copy of Lee and see what you are talking about. In local coordinates since you look like inclusion of a subspace, it makes sense to talk about the same $\epsilon$ in $U$ and $V$. It's true that you may have stretching in directions normal to $i(U)$ in $V$, but on $i(U) \cap V$, it makes sense to talk about the same $\epsilon$ and because of this exactly, $i(U_0)$ is a $k$-slice in $V_0$. $\endgroup$ Commented Jul 23, 2020 at 1:28
  • $\begingroup$ A sort of cleaner way to say this is that the coordinate representation is an isometry from $U$ to $V$ so an open disk of radius $\epsilon$ in $U$ centred at $p$ lies inside an open disk of a higher dimension of the same radius in $V$ centred at $p$, exactly how it looks in the top right disk inside a disk in Fig 5.2. $\endgroup$ Commented Jul 23, 2020 at 1:31
  • $\begingroup$ @OsamaGhani I understand what is going on with the coordinate representations, but I still cannot see how this yields the result. Must an isometric coordinate representation have the desired properties? What if the inverse of the first chart map "stretches" $U_0$, and the second chart map "shrinks" it? $\endgroup$
    – Victor
    Commented Jul 23, 2020 at 2:09

1 Answer 1


I am not sure if this is related to isometric coordinate representation, in fact I don't think we need the notion of metric here. For $U$ open in $S$ and $V$ open in $M$, we have coordinate representation of the inclusion map being $(x^1,\ldots,x^k)\rightarrow (x^1,\ldots,x^k,0,\ldots)$. Since $U$ and $V$ are open, we can choose a $k$ dimensional ball of radius $\epsilon_1$ contained in $U$, and a $n$ dimensional ball of radius $\epsilon_2$ in $V$, centered at the point. We simply choose $\epsilon={\rm min}\{\epsilon_1,\epsilon_2\}$. Let $U_0$ be the $k$ dimensional ball of radius $\epsilon$ which is contained in $U$, and $V_0$ be the $n$ dimensional ball of radius $\epsilon$ which is contained in $V$. By the inclusion map, $U_0$ is mapped to identically the same ball which is contained in $V_0$, because they have the same radius just different dimension, like a disk is contained in a 3D ball. We don't need metric, we just need to know each point in the ball is identified with a point in the manifold.


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