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So I've read Loring Tu's Introduction to Manifolds and now am looking at Guillemin and Pollack's Differential Topology. Tu uses "intrinsic" definitions of smooth manifolds and the like i.e. a certain kind of topological space with a maximal atlas and such. G&P consider manifolds as subsets of $\mathbb{R}^N$ for some $N$ s.t. at each point the manifold is locally diffeomorphic to $\mathbb{R}^k$ for some fixed $k$ (i.e. there exists an open (in the manifold $M$) neighborhood $U$ of a point $x$ that is diffeomorphic to an open set $V$ in $\mathbb{R}^k$; diffeomorphic is a a smooth bijection with a smooth inverse, see below for smooth on non-open subsets of Euclidean space). Whatever, there is some theorem that every manifold can be embedded into Euclidean space (which I haven't seen a proof of but that's besides the point) and anyway the only manifolds I care about I constructed as subsets of some large ambient Euclidean space anyway, and clearly a G&P manifold is a Tu manifold.

I want to understand the correspondence between the definitions for manifolds defined intrinsically and as subsets of Euclidean space. In this question, I'm considering the definition of a smooth map. Let's first go with Tu's definition. Let $M$, $N$ be manifolds and $f: M \to N$ a continuous function. Let $p \in M$. Say $f$ is smooth at $p$ if there are charts $(U, \phi), (V,\psi)$ s.t. $p \in U, f(p) \in V$ and $\psi \circ f \circ \phi^{-1}: \phi(f^{-1}(V) \cap U) \to \mathbb{R}^n$ is smooth. $f$ is smooth if $f$ is smooth at $p$ for every $p \in M$.

Now G&P. Let $X \subseteq \mathbb{R}^N$ for some $N$. Call $f: X \to \mathbb{R}^m$ smooth if for all $x \in X$ there exists an open neighborhood $U\subseteq \mathbb{R}^n$ of $x$ and an extension of $f|_{U \cap X}$ to $U$ that is smooth. Then for $Y \subseteq \mathbb{R}^m$, we have $f: X \to Y$ is smooth if it is smooth as a function $f: X \to \mathbb{R}^m$.

It is easy to verify G&P smooth implies Tu smooth for embedded submanifolds of Euclidean space. How does one show the converse, i.e. supposing a function between two embedded submanifolds of Euclidean space is Tu smooth, why is it G&P smooth. This seems to involve constructing extensions of smooth functions.

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  • $\begingroup$ What is the definition of "locally diffeomorphic to $\mathbb R^k$ for some $k$" in G&P? By the way, it cannot be a point having this property. $\endgroup$
    – Paul Frost
    Jan 17, 2020 at 0:31
  • $\begingroup$ I made some edits to clarify; the wording was bad. Note though that the section you are referring to is primarily for context and not necessarily integral to the proper question. $\endgroup$ Jan 17, 2020 at 0:42
  • $\begingroup$ The easiest way to get local smooth extensions is to apply the immersion theorem to get local coordinates $(x_1,\dots,x_N)$ on $\Bbb R^N$ so that the piece of $X$ looks like $x_{k+1}=\dots=x_N=0$. Given a function $f(x_1,\dots,x_k)$, the obvious extension is $F(x_1,\dots,x_N) = f(x_1,\dots,x_k)$. (Geometrically, we're taking a tubular neighborhood and extending $f$ by letting it be constant, say, on normal cross-sections.) $\endgroup$ Jan 17, 2020 at 5:00
  • $\begingroup$ The defnition in G&P is essential. See my answer. $\endgroup$
    – Paul Frost
    Jan 17, 2020 at 12:02

2 Answers 2

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Your question is essentially covered by my answer to Equivalent definition of a tangent space? However, my answer did not focus on the concept of a smooth map, so let me do it properly here.

First note that property "Tu-smooth at $x \in X$" does not depend on the choice of charts around $x$ and $f(x)$. This is a basic theorem.

G&P define a manifold as a subset $X \subset \mathbb R^N$ if for each $x \in X$ there exist an open neighborhood $U$ in $X$ and a diffeomorphism $\psi : U \to V$ onto an open $V \subset \mathbb R^k$. Here "diffeomorphism" means that $\psi$ is a bijection such that both $\psi, \psi^{-1}$ are "G&M-smooth" in the sense defined in your question.

The first thing we have to do is to endow a G&M-manifold with a differentiable structure in the sense of Tu. As an atlas take the family of all local diffeomorphisms $\psi : U \to V$ as above. Now have a look at the above link. You will see that for each $z \in U$ there exist an open neigborhood $W$ of $z$ in $\mathbb R^N$, $W \cap X \subset U$, and a smooth extension $F : W \to V$ of $\psi \mid_{W \cap X}$.

Now the transition map between charts $\psi : U \to V$ and $\psi' : U' \to V'$ is given by $$\psi' \circ \psi : \psi^{-1}(U \cap U') \to \psi'(U \cap U') .$$ But this map is smooth: We know that $\overline{\psi^{-1}} : U \to \mathbb R^N, \overline{\psi^{-1}}(\zeta) = \psi^{-1}(\zeta)$, is smooth, and that around each $z = \psi(\zeta)$ with $\zeta \in \psi^{-1}(U \cap U')$ we have a smooth extension $F' : W' \to U'$ of $\psi' \mid_{W' \cap X}$ for some open neigborhood $W'$ of $\psi(\zeta)$ in $\mathbb R^N$. We may assume that $W' \subset U \cap U'$. This shows that $\zeta$ has an open neighborhood conatined in $\psi^{-1}(U \cap U')$ on which $\psi' \circ \psi$ agrees with $F \circ \overline{\psi^{-1}}$ which is smooth.

Thus our atlas is smooth and generates a differentiable structure on $X$.

It is then easy to see that the inclusion $j : X \to \mathbb R^N$ is Tu-smooth. In fact, let $\psi : U \to V$ be a chart in our atlas. On $\mathbb R^N$ take the identity as a chart. Then $id \circ j \circ \psi^{-1} = \overline{\psi^{-1}}$ which is smooth.

Now let us verify that Tu-smooth maps are G&P-smooth. Let $f : X \to Y$ be Tu-smooth at $x \in X$. Then also $\bar f = j \circ f : X \to \mathbb R^N$ is Tu-smooth. This means that for any chart $\psi : U \to V$ around $x$ tha map $\bar f \circ \psi^{-1} : V \to \mathbb R^N$ is smooth. There is a smooth extension $F : W \to V$ of $\bar f \mid_{W\cap X}$, where $W$ is some open neigborhood of $x \in \mathbb R^N$. Thus also $ \bar f \circ \psi^{-1} \circ F : W \to \mathbb R^N$ is smooth. Clearly it is an extension of $\bar f \mid_{W\cap X}$.

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  • $\begingroup$ I appreciate the answer, but I was still left confused about some things (in particular the third to last sentence asserts the existence of a smooth extension, but I don't see why one necessarily exists, at least in the context given). I have spent a good chunk of time demonstrating the correspondence between Tu's definitions and G&P's definitions to my own satisfaction; I'm going to attach my write up as a new answer. $\endgroup$ Jan 19, 2020 at 23:20
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I was still confused after looking at Paul's answer and wanted to just work through everything from first principles to make sure I understood. I've written up a somewhat lengthy but very explicit demonstration of the correspondence between Tu's and G&P's definitions of smooth manifolds, smooth maps, tangent spaces, and differentials. I will leave a link to the pdf of the write up here for anyone who might want it:

https://drive.google.com/file/d/1WvfwaKvKLp13Swioqy0ElLb7zLxisi_D/view?usp=sharing

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