Is it true that a hypergeometric function can be analytically continued to a Riemann surface for $\log(z-1)$? It is well known that hypergeometric function $_2F_1(a,b,c;z)$ can be analytically continued to a slit complex plane which is $\mathbb{C}\backslash[1,\infty)$.
What is more, Monodromy theorem on Riemann surface shows that if $X$ is a simply connected Riemann Surface, $a\in X$ and $\phi\in \mathcal{O}_a$ is a function germ which admits an analytic continuation along every curve starting at $a$, then there exists a globally defined holomorphic function $f\in\mathcal{O}(X)$ such that $\rho_a(f)=\phi$, where $\rho_a(f)$ means the germ of $f$ at the point $a$.
Based on the theorem above, it seems that I can analytic continuate the hypergeometric function to the Riemann surface $Y$ for $\log(z-1)$ whose charts map from $\mathbb{C}\backslash\{1\}$ to $\mathbb{C}$, since $Y$ is simply connected and the points on $[1,\infty)$ are branch points. Is it true? Thanks for your explaination.
For easy understanding, I put a figure which describes the shape of $Y$,

 A: In general, no.  Suppose we take a ${}_2F_1(a,b;c;z)$, continue it analytically, starting at $z=0$, into the lower half-plane, cross the real axis above $1$, then continue in the upper half-plane back to $0$.  We could end up with a singularity at $0$.  The differential equation for ${}_2F_1(a,b;c;z)$ has singularities at $0$, $1$, and $\infty$.  But $\log(z-1)$ has singularities only at $1$ and $\infty$.

Example.  The differential equation
$$
4z(1-z)u''(z)+(4-8z)u'(z)-u(z) = 0
\tag1$$
has general solution near $z=0$ given by $C_0 u_0(z)+C_1u_1(z)$,
where
\begin{align}
u_0(z) &= {}_2F_1\left(\frac12,\frac12;1;z\right)
=1+\frac{1}{4}z + \frac{9}{65}z^2+\dots
\\
u_1(z) &= \pi\;{}_2F_1\left(\frac12,\frac12;1;1-z\right) =
\ln\frac{1}{z}+4\ln 2+\dots
\end{align}
Start with the function $u_0$ and analytically continue it along the circle
$$
z = 1-e^{it},\quad 0 \le t < 2\pi .
\tag2$$
When we arrive back at $0$ we are on the branch
$u_0 - \frac{2i}{\pi} u_1$ .  This branch goes to $\infty$ as we approach $0$.  Here is a graph of the imaginary part of the function along that curve:

The red part is the imaginary part of the image of the lower half of the circle $(2)$, starting at $z=0$; the blue part is the imaginary part
of the image of the upper half of the circle, logarithmically approaching $-\infty$ as $t$ approaches $2\pi$.
Summary: the analytic continuation of the hypergeometric function ${}_2F_1\left(\frac12,\frac12;1;z\right)$, going around the point $z=1$, is a different solution of $(1)$ when we arrive back at $0$.
