# Finding holes in a non-simply connected open subset of $\mathbb{C}$

Let $\mathbb{C}^{*} = \mathbb{C} \setminus \{0\}$, and $U \subset \mathbb{C}^{*}$ an open, connected, not simply connected subset. Can we find a holomorphic $f : U \to \mathbb{C}^{*}$ such that $f_{\sharp} : \pi_1(U) \to \pi_1(\mathbb{C}^{*})$ (where I've omitted the choice of basepoint) is nontrivial?

So essentially we want to find a hole of $U$ and translate that hole to the origin (at least, this is the approach that comes to me). If we could find a Jordan curve lying inside $U$ which is homotopically nontrivial, I believe we could do the following: this Jordan curve bounds a (topological) disk. Clearly $U$ cannot contain all points in this disk or else the curve would be homotopically trivial. So take a point in the disk but not in $U$ and consider the translation which sends it to the origin; I believe that should induce something nontrivial on $\pi_1$.

However, I'm not sure if we can always find this Jordan curve. For instance, it doesn't seem to be true that every element of $\pi_1(U)$ can be represented by a Jordan curve (imagine the figure $8$ as an element of $\pi_1(\mathbb{C} \setminus \{0, 1\})$).

• I'm not sure I understand your doubt. You want the map to be nontrivial, that is, there must be at least an element of $\pi_1(U)$ which is not sent to zero via f_#. In the case of the complex plane without two points, just choose one of the loops around a single point to be sent in the non-zero loop in the complex plane without a point. May this work?
– nelv
Sep 28 '15 at 9:44

By my argument in this answer, any loop in $U$ is homotopic (in $U$) to a piecewise linear loop. With a bit of care, we can also arrange that this loop never retraces the same line segment, so it only intersects itself finitely many times (this should be easy to see from the proof). Call such a non-retracing piecewise linear loop good. Define an intersection of a good loop $\gamma:[0,1]\to U$ to be a pair $(s,t)\in[0,1]^2$ such that $s<t$ and $\gamma(s)=\gamma(t)$. Define the complexity of a good loop $\gamma:[0,1]\to U$ to be the number of intersections it has. A good loop has complexity $1$ iff it is a Jordan curve (since $(0,1)$ is always an intersection).

Now suppose $\gamma:[0,1]\to U$ is a non-nullhomotopic good loop which has minimal complexity among all non-nullhomotopic good loops in $U$. Suppose that $\gamma$ is not a Jordan curve; then there is some intersection $(s,t)\neq(0,1)$ of $\gamma$. Then the loop obtained by restricting $\gamma$ to $[s,t]$ is also good and has lower complexity than $\gamma$, so it must be nullhomotopic. Using such a nullhomotopy between $s$ and $t$ and staying fixed outside of $[s,t]$, we obtain a homotopy of $\gamma$ to the loop which is equal to $\gamma$ outside of $[s,t]$ and constant on $[s,t]$. Contracting $[s,t]$ to a point, we then get a new good loop $\gamma'$ which is homotopic to $\gamma$ and has lower complexity than $\gamma$. This contradicts the minimality of $\gamma$.

Thus $\gamma$ is a Jordan curve which is not nullhomotopic in $U$, and by your argument we can get a map $U\to \mathbb{C}^*$ that is nontrivial on $\pi_1$. Note that for this, you only need the fact that a piecewise linear Jordan curve bounds a disk, which is far more elementary than the corresponding fact for arbitrary Jordan curves.

Good textbooks on complex analysis show that an open set $\Omega\subseteq\mathbb C$ is simply connected if and only if every nowhere-zero holomorphic function on $\Omega$ has a holomorphic square root.

So if $\Omega$ is not simply connected, there is a holomorphic function $f:\Omega\to\mathbb C^\times$ which does not have a holomorphic square root. Pick $z_0\in\Omega$, put $w_0=f(z_0)$, and consider the map $f_*:\pi_1(\Omega,z_0)\to\pi_1(\mathbb C^\times,w_0)$. If $f_*$ were trivial, the fact that the map $\exp:\mathbb C\to\mathbb C^\times$ is a universal covering would imply that there is a conttinuous function $g:\Omega\to\mathbb C$ such that $f(z)=\exp g(z)$ for all $z\in\Omega$, and it is easy to see from this that $g$ would have to be holomorphic. But then we the function $\exp\tfrac12g(z)$ would be a square root of $f$, contradicting the way we chose the latter.

• Among the good textbooks on complex analysis are the ones by Remmert, Conway, and Lang. Sep 29 '15 at 4:41
• To provide a bit more background on the theorem you're quoting, it follows from the most commonly given proof of the Riemann mapping theorem, since the only consequence of simple connectedness which that proof uses is the existence of square roots of nowhere zero functions. Sep 29 '15 at 5:06
• Your answer is a perfectly reasonable one and probably the best proof of this fact. I'm accepting the other one because (a) it was posted (slightly) earlier (b) it follows the geometric spirit I hoped for this question Sep 29 '15 at 13:23

You cannot represent every homotopy class with a Jordon curve, but you can always represent one. The construction goes as follows. Take a map $\gamma:I\to U$ with $\gamma(0)=\gamma(1)$ which represents a nontrivial homotopy class. Find an $\epsilon$-ball around $\gamma(I)$ in $U$, and take a polynomial approximation within this neighborhood. The ends don't need to meet up, so we can connect them with a line. This gives us a piecewise polynomial function, which thus intersects itself a finite number of times. If $\gamma_i:S^1\to U$ are all the curves acheived by breaking our closed curve along its self-intersections, then we have $\prod_i \gamma_i=\gamma$, so some closed curve is trivial, but it has no self-intersections by assumption, which gives your desired result.

• Can you include more details? I'm not sure what is meant by polynomial approximation here. It's also not clear how one breaks the curve up into a finite product of closed curves (I don't think that's true even; what does that look like for the figure 8?) Sep 18 '15 at 12:32