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In basic calculus, one partial-differentiate a differentiable function whose domain is an open set or a closed set etc. However how formally this process works?

Here is a reference : definition of winding number, have doubt in definition.

Define $p:\mathbb{C}\rightarrow \mathbb{C}\setminus\{0\}:z\mapsto e^z$.

Then, it is a covering map, hence if $\alpha:[0,1]\rightarrow \mathbb{C}\setminus\{0\}$ is a given path, there is a lift $\tilde{\alpha}$ such that $p\circ \tilde{\alpha}=\alpha$, by Homotopy lifting theorem. This means that $\alpha$ can be decomposed into continuous "length" part and "angle" part.

However, consider a continuous function $f:V\rightarrow \mathbb{C}\setminus\{0\}$ where $V$ is open in $\mathbb{C}$. Saying $f$ can be decomposed into continuous length part and angle part, means that there exists a lift $g$ of $f$. Does this always exist? If so how? If not, why is polar coordiate introduced so uncautiously?

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    $\begingroup$ An open annulus might be hard to lift nicely, but that's why you have expressions like branch cut and main branch. $\endgroup$ – Arthur Dec 1 '14 at 11:57
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Not necessarily, for example if $V=\mathbb C\setminus\{0\}$ itself and $f$ is the identity. Then the argument part can't be chosen continuously.

It should be true if $V$ is required to be simply connected rather than open.

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  • $\begingroup$ I'm confused. I'm sure that the concept of simply connectedness would be never introduced in basic calculus text, but why really why is it introduced extremely incautiously in most texts and why people never care how this process formally works?? $\endgroup$ – Rubertos Dec 1 '14 at 12:04
  • $\begingroup$ It's not clear to me what the "incautiously" you refer to is. Are you saying you have seen books claim (or imply) that the property extends from intervals of the real line to more general sets of complex numbers? I haven't seen that. $\endgroup$ – Henning Makholm Dec 1 '14 at 12:12
  • $\begingroup$ Yes. Some sets are not explicitly named as "open" or "closed", but there are some explicit subsets of $\mathbb{R}^2$ such as ellipse. $\endgroup$ – Rubertos Dec 1 '14 at 12:20
  • $\begingroup$ And I'm trying to figure out how simple connectedness of $V$ implies that.. Would you give me some hints? So far, I have proven the case when $V$ is compact contractible $\endgroup$ – Rubertos Dec 1 '14 at 12:22
  • $\begingroup$ @Rubertos: I would imagine something like: Choose the argument at one particular point $P_0$. At every other point $P$, define the argument by taking some path from $P_0$ to $P$ and applying the $[0,1]$ variant. Because of simple connectedness any two such paths are homotopic, and the argument choice must change continuously; because the possible arguments are discrete this means that any two paths will lead to the same argument at $P$. Actual continuity of this choice is then mostly fiddlework, I think. $\endgroup$ – Henning Makholm Dec 1 '14 at 12:38

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