2

I am a 'lowly physicist' (actually, only a prospective physicist) with little to no background in formal mathematics. Recently I decided it's time to get serious, so I started reading Spivak's 'Comprehensive Introduction to Differential Geometry'. The first chapter, on manifolds, starts off with the definition of a manifold:

Definition: A manifold is a metric space $M$ with the property that, for any point $x\in M$, there exists a neighborhood $U$ of $x$ and some integer $n\geq 0$ such that $U$ is homeomorphic to $\mathbb{R}^n$.

On the third page, Spivak argues that $U$ must be an open set. He has already demonstrated that we can always choose $U$ to be open. He gives, without proof, the 'Invariance of domain' theorem:

Theorem: If $U\subset \mathbb{R}^n$ is open and $f: U\to \mathbb{R}^n$ is one-one and continuous, then $f(U)\subset \mathbb{R}^n$ is open. (it follows that $f(V)$ is open for any open $V\subset U$, so $f^{-1}$ is continuous and $f:U\to f(U)$ is a homeomorphism.)

Now, Spivak states that it is 'an easy exercise' to show that this theorem implies that $U$ from the above definition must be open.

I have tried to look up and understand all the relevant definitions. What I (think I) have understood so far is the following. From the property that $U$ is homeomorphic to $\mathbb{R}^n$ it follows immediately that there is also a homeomorphism (i.e. a continuous and one-one (?) function) from $\mathbb{R}^n$ to $U$. If we denote the homeomorphism from $U$ to $\mathbb{R}^n$ by $f$, it is enough to show that $f(U)$ is open to conclude that $f$ is open by the invariance of domain theorem. However, it is not clear to me that this must be the case. Am I missing a subtlety in a definition? Or is there some nontrivial step here?

Danu
  • 1,772
  • Your statement of the "invariant of domain" theorem looks to have an error. One does not say that a subset of $\mathbb{R}^n$ is one-to-one and continuous. – Lee Mosher May 28 '14 at 15:13
  • A homeomorphism is not only one-to-one, but is also onto (also called surjective). Also, the Invariance of Domain theorem should say "$f$ is a homeomorphism from $U$ to its image $f(U)$" (the map $f: U \rightarrow \mathbb{R}^n$ is not necessarily a homeomorphism). Does this make things a little more clear? – John M May 28 '14 at 15:50
  • @JohnM I am not sure I understand the distinction you're trying to highlight in your penultimate sentence. Could you elaborate? – Danu May 28 '14 at 16:08
  • Considering the $f$ described in the Invariance of Domain theorem, $f: U \rightarrow f(U)$ is a homeomorphism, but $f: U \rightarrow \mathbb{R}^n$ is not a homeomorphism (unless $f(U) = \mathbb{R}^n$). – John M May 28 '14 at 16:36
  • @JohnM Ah, so this is a subtlety applying to the very last part of my statement of the theorem. I'll adjust it accordingly. – Danu May 28 '14 at 16:38

1 Answers1

7

The question is to prove that if $U \subset M$ is homeomorphic to $\mathbb{R}^n$ then $U$ is open. Choose a homeomorphism $f :U \to \mathbb{R}^n$ (yes, homeomorphisms are one-to-one, onto, and they are continuous with continuous inverse; also, to be explicit, the topology on $U$ is the subspace topology, whose open sets are the intersections of $U$ with open subsets of $M$).

Take a point $x \in U$. We can choose an open subset $V_x \subset M$ containing $x$ and a homeomorphism $f_x : V_x \to \mathbb{R}^n$. The set $V_x \cap U$ is open in the subspace topology on $U$. Therefore $f(V_x \cap U) \subset \mathbb{R}^n$ is open, and it contains $f(x)$. There is an open ball $B \subset \mathbb{R}^n$ centered on $f(x)$ such that $B \subset f(V_x \cap U)$. The function $f_x \circ f^{-1}$ restricted to $B$ is one-to-one and continuous, and so by applying the Invariance of Domain Theorem its image $f_x \circ f^{-1}(B)$ is an open subset of $f_x(V_x)=\mathbb{R}^n$. It follows that $f^{-1}(B) = f_x^{-1} \circ f_x \circ f^{-1}(B)$ is an open subset of $V_x$ which is an open subset of $M$. Therefore $f^{-1}(B)$ is an open subset of $M$. Also, by construction $x \in f^{-1}(B) \subset U$. Since $x \in U$ is arbitrary, $U$ is open.

Lee Mosher
  • 120,280
  • It appears the crux of your argument is in the sentence : "The function $f_x∘f^{−1}$ restricted to B is one-to-one and continuous"

    I am not sure how this follows from the previous statements (which I think I do understand). Could you elaborate? Is it correct to say that this is because $f$ is a homeomorphism?

     – Danu May 28 '14 at 16:04
  • I also am not sure whether this argument proves that we can always choose $U$ open, or whether $U$ must be open. In particular, it appears to be very similar to the argument that Spivak presents for the former (which is not what I'm looking for). Could you clarify this issue for me? – Danu May 28 '14 at 16:07
  • 2
    Regarding your first comment, each of $f_x$ and $f$ is a homeomorphism: they are one-to-one, onto, continuous, and have continuous inverses. The same is therefore true of the inverse function $f^{-1}$. The restriction of any homeomorphism to a subset is a homeomorphism from that subset to its image. So $f^{-1}$ restricted to $B$ is a homeomorphism from $B$ to $f^{-1}(B)$. You can now compose that with $f_x$, to get the function $f_x \circ f^{-1}$ restricted to $B$. This is a composition of one-to-one, continuous functions, so it is one-to-one and continuous. – Lee Mosher May 28 '14 at 17:40
  • 2
    Regarding your second comment, all I have assumed is that $U$ is a subset of $M$ homeomorphic to $\mathbb{R}^n$, and what I have proved is that for every $x \in U$ there is an open subset of $M$, namely the subset $f^{-1}(B)$, that contains $x$ and is contained in $U$. So $U$ is a union of open subsets of $M$, implying that $U$ is itself open. – Lee Mosher May 28 '14 at 17:44
  • @LeeMosher This is a little old, but for the benefit of future readers (since this tripped me up): I don't think it's obvious, given Spivak's definition of a manifold, that $V_x$ and $V_y$ are homeomorphic to Euclidean spaces of the same dimension. Indeed this need not be true under Spivak's definition if $x$ and $y$ live in different components of $M$. But I think the trouble could be resolved with a connectedness argument like this one. – Nick Strehlke Oct 06 '14 at 19:33
  • Nitpicking, but in the first sentence of the second paragraph, $x$ should be taken arbitrarily from $U$ rather than from $M$ (the last sentence of the proof makes it clear that the author intended $x \in U$). I tried to edit, but edits have to be 6 or more characters long. – Andrew Estrella Jan 22 '20 at 02:46
  • @AndrewEstrella: After some time I noticed your comment and made the change; I guess authors have more editing authority. – Lee Mosher Oct 21 '22 at 13:08