# Embedded submanifolds satisfy local slice criterion

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 $$\iota(U_0) \subseteq V_0$$.

• 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. Jul 23, 2020 at 0:49
• @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. Jul 23, 2020 at 0:58
• 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 $V_0$. Jul 23, 2020 at 1:28
• 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. Jul 23, 2020 at 1:31
• @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? Jul 23, 2020 at 2:09

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.