I am reading Do Carmo's Differential Geometry of Curves and Surfaces. He defines differentiability as follows:

A continuous map [emphasis mine] $\varphi: V_1 \subset S_1 \to S_2$ of an open set $V_1$ of a regular surface to a regular surface $S_2$ is said to be differentiable at $p \in V_1$ if, given parametrizations $$x_1 : U_1 \subset \mathbb{R}^2 \to S_1 \quad x_2 : U_2 \subset \mathbb{R}^2 \to S_2 $$ with $p \in x_1(U_1)$ and $\varphi(x_1(U_1)) \subset x_2(U_2),$ the map $$x_2^{-1} \circ \varphi \circ x_1:U_1 \to U_2$$ is differentiable at $q = x_1^{-1}(p)$.

I believe that when he says continuous, and when he says that $V_1$ is open, he is referring to the topologies inherited from $\mathbb{R}^3$.

I was wondering why he assumes $\varphi$ to be continuous in the definition. I thought that it must be an easy consequence that $\varphi$ is continuous, if you know that all maps of the form $x_2^{-1} \circ \varphi \circ x_1$ are differentiable (hence continuous).

However, I am starting to think that the continuity requirement is not superfluous. I think that for the definition to be correct, we must do one of two things. Either

  1. Assume $\varphi$ is continuous.
  2. In the definition, change "given parametrizations" to "there exist parametrizations".

If we do neither, then I think there can be very badly behaved, discontinuous functions $\varphi$ where it is impossible to find parametrizations $x_1, x_2$ with $\varphi(x_1(U_1)) \subset x_2(U_2)$. Hence, $\varphi$ would vaccuously be a diffeomorphism, even though it is discontinuous.

However, if we do drop continuity but make change in $2$, I think we can prove that $\varphi$ is continuous, and we get an equivalent definition to do Carmo's.

Is my analysis above correct?


Proposition: Let $\varphi$ be any function from $S_1$ to $S_2$. If for each point $p \in S_1$ there exist parametrizations $$x_1 : U_1 \subset \mathbb{R}^2 \to S_1 \quad x_2 : U_2 \subset \mathbb{R}^2 \to S_2 $$ with $p \in x_1(U_1)$ and $\varphi(x_1(U_1)) \subset x_2(U_2)$ such that the map $$x_2^{-1} \circ \varphi \circ x_1:U_1 \to U_2$$ is differentiable at $q = x_1^{-1}(p),$ then $\varphi$ is a continuous map from $S_1 \to S_2$ (with the topologies inherited from $\mathbb{R}^3$).

$\textit{Proof: }$ Fix $p \in S_1$ and take parametrizations $x_1, x_2,$ and the point $q$, as given in the claim. We will show that $\varphi$ is continuous at $p$.

Let $O_{S_2}$ be any open subset of $S_2$ containing $\varphi(p)$, and let $O_{U_2} = x_2^{-1}(O_{S_2})$, which is open because (by definition of parametrization) $x_2$ is a homeomorphism.

Now $x_2 ^{-1} \circ \varphi \circ x_1$ is differentiable at $q$, so it is continuous at $q$. Since $O_{U_2}$ contains $x_2 ^{-1} \circ \varphi \circ x_1(q)$, there exists an open subset $O_{U_1}$ of $U_1$, containing $q$, such that $x_2 ^{-1} \circ \varphi \circ x_1 (O_{U_1}) \subseteq O_{U_2}.$ Applying $x_2$ to both sides, we get $\varphi \circ x_1 (O_{U_1}) \subseteq x_2(O_{U_2}) = O_{S_2}.$

Finally, let $O_{S_1} = x_1(U_1)$, which is open because $x_1$ is a homeomorphism. Substituting above, we have $\varphi(O_{S_1}) \subseteq O_{S_2}$, which shows that $\varphi$ is continuous at $p$. Since $p$ was arbitrary, $\varphi$ is continuous on $S_1.$

  • $\begingroup$ It is all fine until the penultimate paragraph: where did a homeomorphism come from here? $\endgroup$ Sep 30, 2023 at 22:35
  • $\begingroup$ @MoisheKohan Ah sorry, I just meant continuous, not a homeomorphism. Is it all good now? $\endgroup$
    – user56202
    Oct 1, 2023 at 0:56
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    $\begingroup$ No, it is still not good. Try to write a formal proof of continuity with your proposed definition to see where things go wrong. $\endgroup$ Oct 1, 2023 at 1:03
  • $\begingroup$ @MoisheKohan Thanks for the suggestion. I have written a proof above in an edit, and I cannot see any error or gap. Could you please take a look? $\endgroup$
    – user56202
    Oct 1, 2023 at 1:42
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    $\begingroup$ @MoisheKohan Oh okay, thank you very much discussing this with me. This is indeed one of Carmo's conditions. I reproduced it in my post, and it is on page 75 of his book. Anyway, I learned a lot by writing out everything in detail, thanks again. $\endgroup$
    – user56202
    Oct 1, 2023 at 12:01


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