# To what scope polar coordinate makes sense?

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?

• An open annulus might be hard to lift nicely, but that's why you have expressions like branch cut and main branch. – Arthur Dec 1 '14 at 11:57

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.
• Yes. Some sets are not explicitly named as "open" or "closed", but there are some explicit subsets of $\mathbb{R}^2$ such as ellipse. – Rubertos Dec 1 '14 at 12:20
• 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 – Rubertos Dec 1 '14 at 12:22
• @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. – Henning Makholm Dec 1 '14 at 12:38