# Is there a good reason why continuity assumed in the definition of differentiability in do Carmo?

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?

EDIT:

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 $$S_1.$$

• It is all fine until the penultimate paragraph: where did a homeomorphism come from here? Sep 30, 2023 at 22:35
• @MoisheKohan Ah sorry, I just meant continuous, not a homeomorphism. Is it all good now? Oct 1, 2023 at 0:56
• No, it is still not good. Try to write a formal proof of continuity with your proposed definition to see where things go wrong. Oct 1, 2023 at 1:03
• @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? Oct 1, 2023 at 1:42
• @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. Oct 1, 2023 at 12:01