There are two ways to define a smooth map between manifolds. The 1st way (for example, Lee): $f:M\rightarrow N$ is smooth iff for every $p\in M$ there exist charts $(U,\varphi)$ at $p$ and $(V,\psi)$ at $f(p)$ such that $f(U)\subset V$ and the map $\psi\circ f\circ \varphi^{-1}$ is smooth from $\varphi(U)$ to $\psi(V)$. The 2nd way (for example, O'Neill): $f:M\rightarrow N$ is smooth iff for all charts $(U,\varphi)$ and $(V,\psi)$ the map $\psi\circ f\circ \varphi^{-1}$ is Euclidean smooth (and defined on an open set of $\mathbb{R}^m$).

The problem with the 2nd way is that $f^{-1}(V)$ need not be open, so our domain $\varphi(U\cap f^{-1}(V))$ may not be open. Some authors give additional assumption that $f$ is continuous which make our domain open and solve the problem. I am wondering can we use a 2nd way definition (without $f$ is continuous) to prove conditions from the 1st way definition. Let us notice that the 1st way definition easily shows that $f$ has to be continuous. I don't see how O'Neill solve the problem. Maybe we should work with extended definition of Euclidean smooth function $f:D\rightarrow \mathbb{R}^n$ on non-open domain $D\subset \mathbb{R}^m$, where for every point $x\in D$ there is an open neighborhood $W\subset \mathbb{R}^n$ of $x$ and a smooth map $F:W\rightarrow \mathbb{R}^n$ that agrees with $f$ on $W\cap D$.

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    $\begingroup$ The wording in O'Neill ("defined on an open set of $\mathbb{R}^n$") seems to imply that as a requirement of the definition $\phi(U\cap f^{-1}(V))$ should be open, for whatever reason, for example if $f$ happens to be continuous. $\endgroup$
    – InTransit
    Commented Aug 19, 2014 at 14:42

1 Answer 1


As @InTransit suggested in a comment, O'Neill's phrase "and defined on an open subset of $\mathbb R^m$" seems to take care of the problem. Note that the subset of $\mathbb R^m$ on which $\psi\circ f \circ \varphi^{-1}$ is defined is $(f\circ\varphi^{-1})^{-1}(V) = \varphi(f^{-1}(V)\cap U)$. O'Neill's definition stipulates that this set is open in $\mathbb R^m$. Because $\varphi$ is, in particular, a homeomorphism from an open subset of $M$ to an open subset of $\mathbb R^m$, this implies that $f^{-1}(V)\cap U$ is open in $M$. Knowing this, we can replace $U$ by $U_0 = f^{-1}(V)\cap U$ and $\varphi$ by $\varphi|_{U_0}$, and then smoothness by O'Neill's definition implies smoothness by mine.

If you omit the requirement that the composite map be defined on an open set, but only require, as the OP suggested, that $\psi\circ f \circ \varphi^{-1}$ have a smooth extension to an open neighborhood of each point, then you won't necessarily get continuity. A counterexample (taken from Problem 2-1 in my smooth manifolds book, 2nd ed.) is the function $f\colon\mathbb R\to \mathbb R$ defined by $$ f(x) = \begin{cases} 1, &x\ge 0,\\ 0, &x< 0. \end{cases} $$

  • $\begingroup$ This is a very nice counter example indeed. $\endgroup$ Commented Aug 20, 2014 at 7:51

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