# Spivak and Invariance of Domain

On p.3 of the first volume of Spivak's Comprehensive Introduction to Differential Geometry, he says that it is an "easy exercise" to show that the invariance of domain theorem (if $f:U\subset\mathbb{R}^n\rightarrow\mathbb{R}^n$ is one-to-one and continuous and $U$ is open then $f(U)$ is open) implies that in his definition of a manifold

a metric space $M$ such that every point $x\in M$ has a neighborhood $U$ of $x$ and some integer $n\geq0$ such that $U$ is homeomorphic to $\mathbb{R}^n$,

the neighborhood $U$ in fact must be open.

My question: My proof seems to require a bit of set up, as well as two appeals two the invariance of domain theorem, including once to first prove that the dimension of a manifold is well defined (which Spivak discusses on p.4). In any case, I feel like the argument is longer than what Spivak calls a "complicated little argument" on the previous page. Am I missing something obvious?

Perhaps my real question is whether this kind of comment is to be expected from Spivak, since I am reading this book on my own.

• I wouldn't let perceived contrasts between 'easy exercises' and 'complicated little arguments' bother you, if you can (a) understand the content of the statement and (b) give a proof that makes use of invariance of domain. [You didn't ask, but IMHO both (a) and (b) are nontrivial, and my own definition of "easy" probably wouldn't include them (although my definition of "elementary" would); in my book, if it's easy to write down a bogus proof, then it's not "easy."] Self-study from Spivak (or similarly glib authors) is much improved if you do not read too much into their informal chitchat. – anon Feb 18 '11 at 4:29
• It is not the theorem that is easy. What Spivak says is easy is the proof that $U$ must be open in his definition of manifold. You should assume the "invariance of domain" and USE it to prove the simple thing. – André Caldas Oct 7 '17 at 1:10

I also find the "it is an easy exercise" remarks very frustrating. I think (but am not sure) that I came up with a proof, probably similar to yours. This assumes the fact stated on page 2 "we can always choose the neighborhood $U$ in our definition to be an open neighborhood." I use this fact in choosing $V$ below.

Let $\varphi : U \to \mathbb{R}^n$ be the homeomorphism given in the definition of the Euclidean neighborhood of $x$, and let $z \in U$.

Since $\, z \in M \quad \exists$ an open set $V$ of $M$ with $z \in V$, and a homeomorphism $\theta:V \to \mathbb{R}^n$.

Consider the set $W \equiv V \cap U$. $W$ is open in the relative topology of $U$, so $\varphi (W)$ is open in $\mathbb{R}^n$. For notational convenience, let $\psi \equiv \varphi \mid_{W}$. The map $\;\theta \circ \psi^{-1} :\varphi (W) \to \mathbb{R}^n$ is continuous, so by Invariance of Domain $\; \theta (W)= \theta \circ \psi^{-1}[\varphi (W)] \;$ is open in $\mathbb{R}^n$, which tells us that $W$ is open in $V$.

Since $V$ is open in $M$, $W$ is open in $M$ and we have $z \in W \subset U$, so $U$ is open.

I wouldn't bet my life that what I did is correct, but I think it is. This is my first time attempting to answer a question, so if I've messed up, please be gentle.

Thanks, Dave

• I think is fine (it is useful to me). @LostInMath please do not forget to upvote this answer and accept it if it's useful to you – leo Aug 10 '12 at 3:47
• I have a doubt about whether the value of n is same for both U and V? – user84854 Jul 3 '13 at 7:20