How to interpret multivalued fields in 2D CFT? In the notes I've seen on 2D Conformal Field Theory, we derive the Witt Algebra by considering infinitesimal transformations of the form
\begin{align}
z' &= z + \epsilon z^{n+1}
\end{align}
which supposedly generate all conformal transformations in the plane since such functions are holomorphic, and thus, complex analytic. This induces the transformation $$\varphi(z) \to \varphi(z') =  \varphi(z) + l_n \varphi(z)$$ where $l_n = -z^{n+1} \partial_z$. From here we go on to obtain all of the interesting results in 2D CFT.
I am struggling with one particular aspect of this, which is that even at the infinitesimal level the mapping $z \to z'$ is not bijective: $z'$ will have roots at $z = 0$ and $z = \epsilon^{-1/n}$. This seems problematic because $\varphi(z')$ could now be a multivalued function, which doesn't seem have any physically sensible interpretation. I would understand if this was a pathology that cropped up at some finite $\epsilon$, but it seems to be a characteristic that holds even at the level of the Lie algebra itself.
What is the resolution of this dilemma?
 A: In the quantum CFT, infinitesimal conformal transformations give rise to Ward identities for correlation functions. For transformations that are not bijective, called local conformal transformations, Ward identities involve not only the fields you started with, but also their Virasoro descendants. A descendant such as $L_{-2}\varphi$ does not have a geometrical interpretation, unlike $L_{-1}\varphi = \partial \varphi$.
What you would call a "physically sensible interpretation" would rather be a "naive geometrical interpretation", and probably does not exist.
Ward identities are all you need in practice: in particular, they allow you to compute conformal blocks.
A: You are correct in your suspicion! I will provide a quote from the "yellow book"[1], chapter 5.1.2.

All that we have inferred from Eq. (5.4) ff. is purely local, that is,
we have not imposed the condition that conformal transformations be
defined everywhere and be invertible. Strictly speaking, in order to
form a group, the mappings must be invertible, and must map the whole
plane into itself (more precisely the Riemann sphere, i.e., the
complex plane plus the point at infinity). We must therefore
distinguish global conformal transformations, which satisfy these
requirements, from local conformal transformation, which are not
everywhere well-defined.

Equation (5.4) refers to the Cauchy-Riemann equations. The global conformal transformations are completely parametrized by
$$f(z)=\frac{az+b}{cz+d}\quad\text{ with }ad-bc=1$$
where $a,b,c,d$ are complex numbers. On the Riemann sphere these transformations form a group and they correspond to translations, scalings, dilations and an operation called the special conformal transformation. Below is an animation of the special conformal transformation applied to my cat, which I used for a presentation. The reason it has the form $\frac{P(z)}{Q(z)}$ with $P$ at most linear is precisely the reason you quoted. By arguing that the same must happen at the point at infinity we can reason $Q$ must be at most linear.
In higher dimensions $d>2$ the local conformal transformations are the same as the global ones and we don't have to make a distinction.
So to conclude, in 2D we have global and local conformal transformations where only the global transformations form a proper group. Nevertheless, the local conformal transformations are still immensely useful.
To answer your question more specifically: holomorphic functions are generally not bijective, yet locally they often are. If a function is holomorphic, it will behave linearly for small enough $dz$. In other words
$$f(z+dz)=f(z)+f'(z)dz+\mathcal O(dz^2)\iff \text{f is holomorphic}$$

Bonus, also from the yellow book. We can identify the global conformal transformations with the following complex matrix
$$\begin{pmatrix}a&b\\c&d\end{pmatrix}$$
the reason for this identification is that when you compose two such transformations, the result is the same as multiplying these two matrices together.
The group of 2x2 matrices, $SL(2,\mathbb C)$, is isomorphic to $SO(3,1)$. This is in line with the result in higher dimensions that the conformal group is isomorphic to $SO(d+1,1)$. The 2D conformal group is not so weird after all.

[1]  Philippe Di Francesco, Pierre Mathieu, and
David Sénéchal. Conformal Field Theory. 1st ed New York, USA: Springer-Verlag, 1997. isbn:
9781461222569.
