Whether or not constant multiplicity one implies chart map

Question

I refer to the exercises of Chapter II.4 of Rick Miranda - Algebraic curves and Riemann surfaces.

Question: Can Exercise II.4E help answer the 2nd part of Exercise II.4A (about converse)?

Guess: I actually think Exercise II.4E answers the 2nd part of Exercise II.4A affirmatively.

What I understand:

1. The only difference I see here is that the 2nd part of Exercise II.4A deals with maps that are like $\phi: U \to V$ where $(U, \phi)$ is a chart of some Riemann surface $X$ and $V$ is open in $\mathbb C$ and Exercise II.4E deals with $f: W \to$ (the whole $\mathbb C$ instead of just some open subset $V$ of $\mathbb C$), where $W$ is an open subset of $p$. (I guess $U$ and $W$ either are or maybe assumed to be connected, which might be needed since in this book connectedness is part of definition of Riemann surface.)

2. I'd think to apply Exercise II.4E to say that for $U$ open in $X$ and for $f: U \to V$: consider extending range of $f$ to get $\tilde f: U \to \mathbb C$: I guess $f$ and $\tilde f$ have the same multiplicity at every point $p \in U$. If $f$ or $\tilde f$ has multiplicity 1 at every point $p \in U$, then $f$ or $\tilde f$ is a local coordinate at every point $p \in U$. And then, well, 'local coordinate function' sounds to me the same as 'chart map'.

Answer 1

I think that in the second part of the first question. it is asked whether every function with constant multiplicity $1$ is a chart for the Riemann surface, Let's say $X$? So what if we want to consider this map as a chart, then it should be a homeomorphism. To see that this claim is not always true. consider the map $e^{z}$ from $\Bbb C$ to $\Bbb C$. around every point the Taylor series of this function starts with a non-constant term of degree $1$. So the multiplicity of the function is $1$ at each point. but $e^{z}$, clearly, is not a chart for $\Bbb C$.

On the other part E is saying that if a function $f$ is can locally be used a chart map around a point $p$ where it is holomorphic of multiplicity $1$. which is true. because around that point the function behaves like the function $z$.