If $S_1, S_2 \subset \Bbb R^3$ are two smooth surfaces, then what is the formal definition of a smooth map from $S_1$ to $S_2$?

I am studying from Pressley's EDG, and the definition is given only in the case where each of $S_1$ and $S_2$ has an atlas containing a single surface patch. Namely, if $\sigma_i$ denotes the surface patch corresponding to $S_i$, $f$ is called smooth if $\sigma_2 ^{-1} \circ f \circ \sigma_1$ is. How does this definition generalize? I know that it has to do with the property of smoothness, as given in the definition, being invariant under reparameterization of patches, but I do not understand how.

Precisely, I am looking for a completely formal definition of a smooth function $f:S_1 \to S_2$ where $S_1$ and $S_2$ do not necessarily have a singleton atlas.

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    $\begingroup$ The condition is just that the (possibly empty) function $\tau_2^{-1} \circ f \circ \tau_1$ (which is between open subsets of $\Bbb R^2$) is smooth for every chart $\tau_1$ in the altas on $S_1$ and chart $\tau_2$ in the atlas on $S_2$. $\endgroup$ Apr 27, 2016 at 9:54
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    $\begingroup$ I mean neither. It is possible to choose surfaces $S_1, S_2$ and atlases $\mathcal A_1, \mathcal B_1$ on $S_1$ and $\mathcal A_2, \mathcal B_2$ on $S_2$ such that $f$ is smooth with respect to the atlases $\mathcal A_1$ and $\mathcal B_1$ but not smooth with respect to the atlases $\mathcal A_2$ and $\mathcal B_2$. Thus, to talk about smoothness in the first place, we need to fix smooth atlases on $S_1$ and $S_2$. $\endgroup$ Apr 27, 2016 at 10:17
  • $\begingroup$ @Travis I realized it and I removed my comment just before you posted your reply. I apologize for that. $\endgroup$
    – user258700
    Apr 27, 2016 at 10:27
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    $\begingroup$ No worries. I'll leave my remark there for any future readers, because this point is, I think, both nonobvious and potentially confusing when you first learn it. $\endgroup$ Apr 27, 2016 at 10:32

1 Answer 1


Don't forget that differentiability is a local property. We say that $f: S_1 \to S_2$ is differentiable at $p \in S_1$ if there is a parametrization $\sigma_1: U_1\subseteq \Bbb R^2 \to V_1 \subseteq S_1$ around $p$ and a parametrization $\sigma_2: U_2 \subseteq \Bbb R^2 \to V_2 \subseteq S_2$ around $f(p)$ such that $\sigma_2^{-1}\circ f \circ\sigma_1$ with the domain appropriately restricted is differentiable at $\sigma_1^{-1}(p)$. There are no worries about we having or not singleton atlases - by definition of surface there will always be at least one parametrization around each point. And as you noted yourself, this does not depend on the choice of parametrization, so we're in business.

Also, do note that we're not using that $S_1,S_2\subseteq \Bbb R^3$ here. Meaning that if you write $U_1 \subseteq \Bbb R^n$ and $U_2 \subseteq \Bbb R^m$ instead, we actually have the general definition of a differentiable map between smooth manifolds.

  • $\begingroup$ How is this response different from mine which got downvoted? If you already have surfaces in $\mathbb{R}^n$ there is no need to parametrize since you can just restrict the one chart atlas $(\mathbb{R}^n, id)$ to these surfaces. Then smoothness of the derivative is exactly what we used to from calculus. $\endgroup$ Apr 27, 2016 at 10:00
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    $\begingroup$ You started saying that $S_1$ and $S_2$ are vector spaces. They are not. You added some stuff later but did not correct the offending bit. $\endgroup$
    – Ivo Terek
    Apr 27, 2016 at 10:02
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    $\begingroup$ Oh geez. I meant vector space, without the ''s'' in reference to $\mathbb{R}^3$. $\endgroup$ Apr 27, 2016 at 10:03
  • $\begingroup$ @FaraadArmwood There were a few issues with your answer ("vector space", unclear wording throughout), but most (all?) of the downvotes it accumulated were before you completely rewrote it, when it didn't address the question as asked. $\endgroup$ Apr 27, 2016 at 10:22
  • $\begingroup$ @Travis: what I was hinting at was the fact that differentiation wouldn't make sense if you're just given a surface. Then I started to talk about parametrize with respect to local coordinates, but whatever, I'll just wait until it's finalized next time. $\endgroup$ Apr 27, 2016 at 10:27

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