0708-1300/Errata to Bredon's Book

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There is a counterexample to the inverse implication in Problem 1, p. 71.

Let $X=\mathbb {R}$ be endowed with the ordinary topology (thus, it is Hausdorff and second countable). Let $U$ be an arbitrary connected open set in $X$ (that is, an interval). Let $F_{X}(U)$ consists of all functions identically equal to constant. If $U$ is an arbitrary open set, then by theorem on structure of open sets in $\mathbb {R}$ it is a union of countably many open intervals. We define $F_{X}(U)$ to be the set of all real-valued functions which are constant on open intervals forming $U$. The family $F=\{F_{X}(U):U{\mbox{ is open in }}X\}$ forms a functional structure, as one can check. Furthermore, it satisfies the hypothesis of the theorem: every point $x\in X$ has a neighborhood (we take an open interval containing $x$) such that there exists a function $f\in F_{X}(U)$ (we define it to be identically equal to $1$) such that a function $g:U\to \mathbb {R}$ is in $F_{X}(U)$ (it is identically equal to a constant by our definition) if and only if there exists a smooth function $h$ such that $g=h\circ f$ (if $g$ is given, then we define $h(x)=g$ for all $x$, if $f$ is given, then we take arbitrary smooth $h:\mathbb {R} \to \mathbb {R}$, since $h\circ f$ is identically equal to constant and, thus, is in $F_{X}(U)$). Clearly, $(X,F_{X})$ is not a smooth manifold. Even taking $X$ as any $T_{2}$ second countable topological space with the functional structure of constant functions will do the work.

Adding to the statement of the problem that the $F=(f_{1},\ldots ,f_{n})$ function is invertible we get a correct theorem. Maybe other weakening of this condition works.