# 0708-1300/Errata to Bredon's Book

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Problem 1, p. 71.

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

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

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

Problem 4, p. 88.

Last line of problem 4 says "Also show that XY itself is not a vector field." and should say "Also show that XY itself is not always a vector field." There are trivial examples in which XY is a vector field. For example if X is identically zero. There are non-trivial examples too but lets give them after the due day of Homework III because I'm sure you will enjoy finding those examples by your self.