I've been reading Bredon's Topology and Geometry recently; what an excellent book! He defines smooth manifolds in two distinct ways and then shows they are in fact equivalent. The "non-standard" definition is in terms of some sheaf-like "functional structure $F_X$ " on the underlying space $X$, satisfying the following properties: for every open set $U \subset X,$ we have

  1. $F_X(U)$ is a subalgebra of the algebra of continuous real-valued functions on $U$;
  2. $F_X(U)$ contains all constant functions;
  3. $V \subset U, f \in F_X(U) \implies f|_V \in F_X(V)$;
  4. $U = \bigcup U_{\alpha}$ and $f|_{U_{\alpha}} \in F_X(U_{\alpha})$ for all $\alpha \implies f \in F_X(U).$

A morphism of functionally structured spaces $(X,F_X) \rightarrow (Y,F_Y)$ is a map $\phi:X \rightarrow Y$ such that $f \mapsto f \circ \phi$ carries $F_Y(U)$ into $F_X(\phi^{-1}(U))$.

Then a smooth $n$-manifold is a second countable, functionally structured, Hausdorff space $(M^n,F)$ which is locally isomorphic to $(\mathbb{R}^n,C^{\infty}).$

My question: to familiarize myself with the definition I have attempted the following exercise:

Show that a second countable Hausdorff space $X$ with a functional structure $F$ is an $n$-manifold $\iff$ every point in $X$ has a neighborhood $U$ such that there are functions $f_1,\ldots,f_n \in F(U)$ such that a real-valued function $g$ on $U$ is in $F(U) \iff$ there exists a smooth function $h(x_1,\ldots,x_n)$ of $n$ real variables such that $g(p) = h(f_1(p),\ldots,f_n(p))$ for every $p \in U.$

The only part that I haven't been able to complete is the "$\Longleftarrow$" direction. That is, given the $n$ "coordinate functions" $f_i$, and given a point $x \in X$ and a neighborhood $U \ni x$ I, have constructed a morphism $\phi:(U,F_U) \rightarrow (\phi(U),C^{\infty})$ via $\phi(x) = (f_1(x),\ldots,f_n(x)).$ But for the life of me, I don't see how I could show that this is actually an isomorphism.

Any hint toward the answer would be greatly appreciated!

  • $\begingroup$ possible duplicate of Functionally structured spaces and manifolds $\endgroup$ – Zhen Lin Aug 10 '13 at 23:16
  • $\begingroup$ @ZhenLin Oh, my bad. If I am reading correctly, does this mean that the problem as stated above is false? $\endgroup$ – Alex Provost Aug 11 '13 at 16:03
  • $\begingroup$ @Alex P. Yes, it is false. I provide an asnwer below. $\endgroup$ – John Aug 22 '13 at 18:32

The implication $\Leftarrow$ you are considering is false. The functionally structured space $(\mathbb{R},F)$ I defined here provides a counterexample. It is not an smooth manifold yet it verifies the property you mention. For every $x\in\mathbb{R}$ we can take any open interval $I$ containing $x$, and let $f_{1}$ be any function in $F(I)$ (recall that $F(I)$ consists only of constant functions).

If $g\in F(I)$ and we let $h=g$ we have that $h$ is smooth of 1 real variable such that $g(y)=h(f_{1}(y))$ for all $y\in I$.

On the other hand, suppose that there is a smooth function of 1 real variable $h$ such that for a continuous $g:I\rightarrow\mathbb{R}$ we have $g(y)=h(f_{1}(y))$ for all $y\in I$. Then since $f_{1}$ is constant, say $f_{1}\equiv c\in\mathbb{R}$, we get $g(y)=h(c)$ for all $y\in I$, i.e., $g\in F(I)$.

If you add the additional hypothesis that $\phi$ is locally invertible then the implication $\Leftarrow$ is true. Other hypotheses may also work.

Othe links to problems on that section of Bredon's book are this, this and this.

  • $\begingroup$ Thanks, John. Would you mind explaining why your counterexample works (more precisely, why the hypotheses are satisfied)? I assume you let $f = id$, as I see no other plausible candidate. But then it's not true that $g(p) = h(p)$ for some smooth $h$ implies that $g$ is locally constant. $\endgroup$ – Alex Provost Aug 23 '13 at 15:49
  • $\begingroup$ @ Alex P. I edited my answer. I hope it helps. $f_{1}=id$ will not work. Just let $f_{1}$ be any constant function. $\endgroup$ – John Aug 24 '13 at 1:21
  • $\begingroup$ Ah, excellent! I don't know why I didn't think of constant functions. Thank you very much, John. $\endgroup$ – Alex Provost Aug 27 '13 at 14:20
  • $\begingroup$ No problem. I edited the question and added some links to questions related to the problems of the section you are reading. I am myself stuck on problem 5 p.71 (see the last link I added at the end of my answer) $\endgroup$ – John Aug 27 '13 at 14:25

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