Problem in my textbook says:
Show that the function $\frac{1}{z}$ could not be analytically continued to along the segment $[-1, 1]$.
So, basically I need to show that there is no chart-covering $(U_i, f_i)$, $i=1,...,k$ of the segment $[-1, 1]$ with $f_i: U_i \to \mathbb{C}$ holomorphic and $$ f_i|_{U_i \cap U_{i+1}} = f_{i+1}|_{U_i \cap U_{i+1}} \\ f_1=1/z \\ $$
My intuition says that there should be problem at $z=0$, but I can't see why.
I think that if we restrict the problem to existence of only $2$ such charts $(U_1, f_1=1/z)$ and $(U_2, f_2)$ and figure out the problem, then we're pretty much done.
But I really can't understand why there could not be such a function $f_2$ which agree with $f_1=1/z$ on $U_1 \cap U_2$, and is holomorphic in $U_2 \ni 0$.
Any ideas?
I asked my professor about the problem.
Hint: Function $1/z$ can be continued to all $\mathbb{C} \backslash \{0\}$. What can be analytic continuation of germ $1/z$ at $0$ then?
Solution: Consider the map $$\tilde{\gamma}: [-1, 1]\to \mathcal{O}$$ of analytic continuation along the segment path $$\gamma: [-1, 1]\to \mathbb{C} \\ t\mapsto t$$ For every $\delta<0$ the result $\tilde{\gamma}(\delta)$ of analytic continuation along a segment $[-1, \delta]$ should be equal to the germ of $1/z$ at that point by the uniqueness of analytic continuation.
Then, by continuity of $\tilde{\gamma}$ we have that $\tilde{\gamma}(0)=1/z$. But this obviously can not be the case since $1/z$ is not analytic at $z=0$.