Last days I have been following an online course about modular functions from Richard Borcherds which I do recommend a lot for anyone interested on the topic. At some point, he uses part of the theory of modular functions in order to provide a proof for Picard Little Theorem in Complex Analysis (from minute 10 in the linked video).
At that point, he has already defined the so-called modular function $j$ from the upper half plane $H$ to $\mathbb{C}$, and he has studied several properties involving it. In particular, he has shown that whenever $j'(\tau)=0$ we have $j(\tau)=0$ or $j(\tau)=1728$. In order to prove Picard, he makes the following argument (or at least, this is what I have understood):
Suppose that we have an entire function $f:\mathbb{C} \to \mathbb{C} \setminus \{0,1728\}$. Then the multivalued function $j^{-1}(f(z))$ avoids the bad points where $j'$ vanishes, so we can select a single valued function from this multivalued function that maps $\mathbb{C}$ into $H$.
However, it is not very clear to me why is this possible. From references, I have seen many mentions to monodromy theorem (but explanations are not very detailed), and it is still not very clear to me how this tool solves the trouble. If we use this statement of Monodromy Theorem…
Let $U$ be an open disk in the complex plane centered at a point $P$ and $f:U\to \mathbb {C}$ be a complex-analytic function. If $W$ is an open simply-connected set containing $U$, and it is possible to perform an analytic continuation of $f$ on any curve contained in $W$ which starts at $P$, then $f$ admits a direct analytic continuation to $W$, meaning that there exists a complex-analytic function $g:W\to \mathbb {C}$ whose restriction to $U$ is $f$.
…, since $j'$ does not vanish, I know that I can make a single-valued local definition for $j^{-1}(f(z))$ near any $z_0$ in $\mathbb{C}$ (but, with an a priori different definition for each $z_0$). Thus, if I take a local univalued definition for $j^{-1}(f(z))$ near $0$ valid in $U$, and I consider a path $\gamma:[0,1] \to \mathbb{C}$ from $0$ to an $w \in \mathbb{C}$, I need to show that I can perform analytic continuation until $w$ along the path.
Of course, I know that at each point $\gamma(t_1)$ of the path I have a local univalued expression for $j^{-1} \circ f \circ \gamma$ valid on an open set $U_{t} \subset [0,1]$ containing $t_1$ (since $j'$ does not vanish at $f(\gamma(t_1))$), but how do I ensure that if $U_{t_1}$ intersects $U_{t_2}$ the two definitions for $j^{-1} \circ f \circ \gamma$ can be chosen to be coherent? For instance, imagine that $j^{-1} \circ f \circ \gamma$ is analytic and univalued until $t<1/2$, but $U_t$ does not contain $1/2$ for any $t<1/2$. What does ensure that I can take a definition for $j^{-1} \circ f \circ \gamma$ on $U_{1/2}$ that matches the previous one? What ensures that $j^{-1} \circ f \circ \gamma$ has limit from the left at $t=1/2$?
Best Answer
To prove this you need to know more than just the properties of $j$ that you have stated so far. You need to know that $j$ is a covering map away from its branch points. That is, if $S=j^{-1}(\{0,1728\})$ then the restriction $j:H\setminus S\to \mathbb{C}\setminus\{0,1728\}$ is a covering map. This means that for each $a\in\mathbb{C}\setminus\{0,1728\}$ there is an open neighborhood $U$ of $a$ such that $j^{-1}(U)$ can be written as a disjoint union of open sets each of which $j$ maps bijectively to $U$.
Knowing that $j'\neq 0$ gives a weak version of this: it tells you that for each $b$ such that $j(b)=a$ there is an open neighborhood $U$ of $a$ and an open neighborhood $V$ of $b$ such that $j$ maps $V$ bijectively to $a$. The covering space property is stronger, though: it says that the $U$ involved here can be chosen to be the same for every single $b$ such that $j(b)=a$. This can be deduced from the symmetry properties of $j$: any two $b,b'$ such that $j(b)=j(b')=a$ are related by a Möbius transformation $g\in SL_2(\mathbb{Z})$ and so if you find a $V$ around $b$ that $j$ maps bijectively to $U$, then $g(V)$ will be a neighborhood of $b'$ that $j$ maps bijectively to $U$.
Once you know that $j$ is a covering map, the existence of a single-valued "$j^{-1}\circ f$" follows from the general theory of covering spaces. Specifically, since the domain of $f$ is simply connected and locally path-connected, there is a continuous map $g:\mathbb{C}\to H\setminus S$ such that $j\circ g=f$. But then, since $j$ is locally a biholomorphism and $f$ is holomorphic, this implies $g$ is automatically holomorphic (since it is locally a composition $j^{-1}\circ f$ of two holomorphic functions).
In terms of your attempted use of the monodromy theorem, the covering space property lets you avoid the difficulty you are running into with a compactness argument. For each $t\in[0,1]$, fix an open disk $U_t$ around $\gamma(t)$ such that $j^{-1}(f(U_t))$ can be written as a disjoint union of open sets each of which $j$ maps bijectively to $f(U_t)$. Let $I_t$ be an open interval around $t$ such that $\gamma(I_t)\subseteq U_t$. By compactness of $[0,1]$, it is covered by finitely many of these open intervals $I_{t_1},\dots,I_{t_n}$, which you can list in order so that consecutive ones overlap. Pick one of the open sets $V_{t_1}\subset j^{-1}(f(U_{t_1}))$ that $j$ maps bijectively to $f(U_{t_1})$, and define $j^{-1}\circ f$ on $U_{t_1}$ using the inverse of $j$ that maps $f(U_{t_1})$ to $V_{t_1}$. Now pick $a\in U_{t_2}\cap U_{t_1}$, and take the open set $V_{t_2}\subset f(U_{t_2})$ that $j$ maps bijectively to $f(U_{t_2})$ which contains our chosen value of $j^{-1}\circ f(a)$ (which we chose before since $a\in U_{t_1}$). We can then define $j^{-1}\circ f$ on $U_{t_2}$ using this $V_{t_2}$, and this will agree with the definition we made on $U_{t_1}$ since they agree at the point $a$ in the overlap (and the overlap $U_{t_1}\cap U_{t_2}$ is connected, being an intersection of two disks). Similarly, you can then pick a definition of $j^{-1}\circ f$ on $U_{t_3}$ that agrees with the one from $U_{t_2}$ on their overlap, and so on to get an analytic continuation of $j^{-1}\circ f$ along $\gamma$.