2

My book is An Introduction to Manifolds by Loring W. Tu.

From Wikipedia: Local diffeomorphism:

For $X$ and $Y$ differentiable manifolds. A function $f:X\to Y$, is a local diffeomorphism, if for each point x in X, there exists an open set $U$ containing $x$, such that $f(U)$ is open in Y and $f|_{U}:U\to f(U)$, is a diffeomorphism.

(I guess the "$f|_U$" is not $f|_U:U \to Y$ but rather $f|_U$ with restricted range $\tilde\{f|_U\}: U \to f(U)$.)

For $X$ and $Y$ instead smooth, and not merely differentiable manifolds, is the assumption that $f(U)$ is open in $Y$ actually redundant?

  • I think we must have $f(U)$ at least a smooth (embedded or regular) submanifold of $Y$ since it wouldn't make sense for $\tilde\{f|_U\}$ to be a diffeomorphism otherwise. Also, I guess we must have $X$, $U$ and $f(U)$ to be of the same dimension.

  • I think that $f(U)$ is open in $Y$ follows from smooth invariance of domain given in Remark 22.5 on Theorem 22.3, which relies on Definition 22.1.

  • Then again I think smooth invariance of domain is not applicable because $X$ and $Y$ are not given as the same dimension. Either I'm missing something or Tu has a different definition of local diffeomorphism (defined in Section 6.7 and further described in Remark 8.12). For example, Wikipedia's local diffeomorphisms are open maps. I'm not sure Tu's local diffeomorphisms are too. Update: I think it's a different definition. See here.

  • I just noticed that Theorem 6.26 and Remark 8.12 have "same dimension". I think Tu's definition is the same as Wikipedia's assuming smooth manifolds of the same dimension.

  • 1
    "We must have $M$,$U$ and $f(U)$ to be of the same dimension". Don't you mean $X$ instead of $M$? – Paulo Mourão Jun 21 '19 at 12:34
  • @PauloMourão No, I do. Thanks! –  Jun 21 '19 at 12:36
  • 1
    By "can we remove", are you asking if the definition would remain the same? i.e., if the assumption is redundant? Or are you asking why isn't the definition changed to allow for non-open images? – Paulo Mourão Jun 21 '19 at 12:46
  • @PauloMourão Redundant. I'll edit actually. Thanks. –  Jun 21 '19 at 12:49

1 Answers1

0

No, it is not redundant. Any embedding is a diffeomorphism onto its image (so you could just take $U=X$ for every point in this case), which, in general, is not an open set in the codomain.

Now, if you assume $\dim X=\dim Y$, with $f(U)$ and embedded submanifold of $Y$, then, as you said, $\dim f(U)=\dim X$ and so you get $\dim f(U)=\dim Y$. You can see here that this does indeed imply that $f(U)$ is open in $Y$. So the assumptions would be equivalent but, honestly, I find the condition of $f(U)$ being open much more intuitive in this case.

I hope this helps (with Question $1$ at least).

Paulo Mourão
  • 1,774
  • Thanks! So to clarify for smooth manifolds but of possibly different dimension, Tu has a different definition of local diffeomorphism from wikipedia? –  Jun 21 '19 at 13:18
  • Also while this is mindblowing, can we just use smooth invariance instead? Or is it actually the same proof? –  Jun 21 '19 at 13:22
  • 1
    Yes, you can. In fact, the result that I mentioned is just that remark $2.25$ that you linked. As for the definition in your book, I think he is assuming the manifolds to have the same dimension, so the definitions match. Even though it is not extremely clear in the first paragraph, he specifically says it from then on, and it's what makes sense, I think. – Paulo Mourão Jun 21 '19 at 13:50
  • Okay I'll have to email Prof Tu again. Thanks. –  Jun 21 '19 at 13:53
  • By the way just to clarify, in this definition of local diffeomorphism, adding any 1 of these 4 conditions is equivalent to adding any other? Condition 1. Each $f(U)$ is open in $M$ Condition 2. $N$ and $M$ have the same dimension. Condition 3. $f(N)$ is open. Condition 4. $f(N)$ is a smooth manifold, and each $f(U)$ is open in $f(N)$, and –  Jun 21 '19 at 13:57
  • 1
    I believe you're always supposed to consider $f(N)$ as an embedded submanifold and, in that case, the first $3$ are equivalent. $4$ isn't, that one is true for general embeddings, since they're diffeomorphisms (and in particular homeomorphisms) onto their images. – Paulo Mourão Jun 21 '19 at 14:06
  • I can't remember what I was thinking for Condition 4 but rereading my comment, I think it's obvious Condition 4 is not equivalent to the others. Thanks! –  Jun 26 '19 at 08:04
  • Wait if you don't consider $f(N)$ an embedded submanifold, then why wouldn't Conditions 1-3 be equivalent? Of course open subset implies embedded submanifold so you don't need to assume it for Condition 3, so I guess the embedded submanifold assumption is for Conditions 1 and 2. –  Jun 26 '19 at 08:09
  • Actually I think Condition 1 and 3 are equivalent imply $f(N)$ is an embedded submanifold. (I'm assuming your embedded submanifold is the same as the one of Tu, who says embedded submanifold is equivalent to regular submanifold) –  Jun 26 '19 at 08:11
  • 1
    The thing about not assuming anything about $f(N)$ being an embedded submanifold and stuff like that is that what you're claiming doesn't make much sense. In the definition that you linked, you end with the claim that $f:U\to f(U)$ is a diffeomorphism. What is the smooth structure on $f(U)$ for you to make that claim? Is $f(U)$ an embedded submanifold of $N$ or what structure are you taking? See what I mean? The claim becomes unclear, in my opinion. – Paulo Mourão Jun 26 '19 at 08:41
  • 1
    Never mind $f(N)$ being an embedded submanifold. I should have said that you're always supposed to consider the $f(U)$'s embedded submanifolds, not $f(N)$. This is because, again, you need some smooth structure on $f(U)$ to claim the diffeomorphism that you want. (Of course that you could assume $f(N)$ to be an embedded submanifold and $f(U)$ to be open in $f(N)$, which is probably what I was thinking, I don't know). – Paulo Mourão Jun 26 '19 at 08:53
  • 1
    Your condition $1$ is, for me, the most natural way of doing things (as it defines a smooth structure on $f(U)$), it's where you're getting at anyways (your aim ends up always being to prove this anyways), and it makes things clearer from the start. Condition $2$ isn't gonna cut it by itself unless you define a smooth structure on $f(U)$ (so, again, I should have told you to assume $f(U)$ was an embedded submanifold, not $f(N)$). Otherwise, how can you claim that diffeomorphism? You can't. Same thing goes for condition $3$. – Paulo Mourão Jun 26 '19 at 08:53
  • Oh right, good point about the $f(U)$. Thanks! Actually, later in my "to ask list", I have an item that is a related question: "How can $f: A \to f(A)$ be a diffeomorphism or local diffeomorphism if $f(A)$ is not a manifold?" I guess that takes care of that. In that case, what do you say to this comment which says "$F(N)$ needs not be a manifold"? –  Jun 26 '19 at 10:02
  • 1
    $f:N\to M$ can only be a smooth map with respect to smooth structures on $N$ and $M$, but this doesn't mean you have a smooth structure on $f(N)$, so it might not make sense to claim that $f:N\to f(N)$ is a smooth map, even if $f:N\to M$ is smooth. – Paulo Mourão Jun 26 '19 at 10:21
  • Thanks! So you disagree with lEm please? –  Jun 27 '19 at 03:54
  • 1
    No, I agree with him. His comment makes sense, he's just saying what I said: you might not have a smooth structure on the image. So in particular, you should be careful with "restricting ranges" like you did in that comment section, it might not make sense (i.e. the map might not be smooth) – Paulo Mourão Jun 27 '19 at 10:05
  • 1
    I also believe he meant that if f(N) is an embedded submanifold, then you can restrict the range and you get a local diffeomorphism when you do so – Paulo Mourão Jun 27 '19 at 10:13
  • Okay thanks. I might come back here if lEm replies. –  Jul 21 '19 at 10:19
  • Paulo Mourão, is the definition that Tu gave originally (the wrong one), actually the definition of a local embedding? –  Jul 24 '19 at 08:54
  • Update: I ask about my previous comment here –  Jul 24 '19 at 09:16