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enter image description here enter image description here Above is the proof of the uniqueness of the covariant derivative given by Do Carmo. My question is that we only know that locally $V=\sum_jv^jX_j$, then How can I jump to the conclusion that $\frac{DV}{dt}=\frac{D\sum_jv^jX_j}{dt}$?

My guess here is by doing this Do Carmo does not actually consider the actual curve $c$, he considers a curve $c'$ whose domain is an open subset of $c$, then $V$ is now viewed as a vector field along $c'$, then the equality makes sense. But from here, how should I prove that my actual $V$ along the original $c$ has covariant derivative that agrees locally with the covariant derivative of $V$ along $c'$ ?

Maybe another way to formalize my question is: Suppose that $c: I \to M$, $c' :I' \to M$ are curves where $I' \subset I$ and they are both open. $V$ is a vector field along curve $c$, and $V'$ is the restriction of $V$ on $I'$ then should $\frac{DV}{dt} = \frac{DV'}{dt}$ agree on every point of $I'$?

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    I guess it is because they are tensor field, if the equation is right in every local , then it will be right global, because the equation of tensor is independent of coordinate. – 2017-01-16

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Do Carmo is defining covariant derivative here, not proving anything. Another way to say this is the following. Fix a chart such that $V$ has the given expression. Then define $\frac{DV}{dt}$ by equation (1). Then check this is well defined, meaning that if you use some other chart the resulting vector field remains unchanged. The last sentence is explained in Do Carmo between the proposition and the proof which you have omitted in the question..

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    I think he does intend to prove the uniqueness using that – 2017-01-16
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    His claim is, once you show it's well defined, uniqueness follows. – 2017-01-16
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    Let me say it in a different way. First you define the covariant derivative using (1). Then you show it is well defined. Then you prove that if it satisfy the properties then it is unique. Do Carmo did it in exactly opposite order. – 2017-01-16
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    Covariant derivative of a vector field $Y$ depends on the value of $Y$ locally. These expressions are local. – 2017-01-16