Let $$ F(U) = \left\{ \mbox{ all complex analytic functions } f \mbox{ on } U \mid z \frac{df}{dz}=1 \right\}$$ for any domain $U$ in $\mathbb{C}$.
I want to show that:
- $F$ is a sheaf.
- The stalk of $F$ at $0$ is empty.
- The stalk at any other point is non-canonically isomorphic to $\mathbb{C}$.
I know that the presheaf of continuous function is a sheaf (as they have the gluing property over open coverings), so also analytic functions (which are $C^\infty$) are sheaf. However, there are two problems is: a function that satisfy $d \frac{df}{dz}=1$ actually is $$\frac{df}{dz} = \frac{1}{z}$$ which is not analytic in 0. Moreover, is not the differential equation $\frac{df}{dz}=\frac{1}{z}$ is unique up to a constant - the family of solution is $f(z) = \ln(z) + c$ with $c \in \mathbb{C}$? Then it follows that $F$ is a sheaf with stalk $$\{f(z) = \ln(z) + c \mid c \in \mathbb{C} \} \cong \mathbb{C}$$ and empty stalk at 0 since there is no analytic function such that $0 \cdot \frac{df}{dt} = 1$. Are my consideration here are correct?
Added in edit:
We can define $\ln(z)$ as the extension of $$ \ln(x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}(x-1)^n $$ to the complex number. Of course one needs to check where it converges and well-defined, and then arise the issue of branch cuts. I am no expert on complex analysis but my intuition says that $\frac{df}{dz}=\frac{1}{z}$ has a family of solutions differ by constants. Is this correct?
I think the point of this exercise is to arrive at the concept of complex logarithm from the sheaf direction, not to fall back onto the power series definition. I suggest not using notation $\ln$, or any prior knowledge of logarithms, in your solution.
The stalk is empty over $0$ because $z=0$ makes it impossible for $zf'(z)=1$ to hold.
Next, fix a point $z_0\ne 0$. Let $(f,\Omega )$ be a function element that satisfies $zf'=1$ in $\Omega$. (Here $\Omega$ is a neighborhood of $z_0$ that does not include $0$). Then for every $c\in\mathbb C$ we also have $z(f+c)'=1$. In particular, we can ensure $f(z_0)=0$ by subtracting a constant from $f$. If $(g,\Omega)$ satisfies $zf'=1$, then $z(f-g)'=0$ which implies $g=f+c$. Thus, we have the isomorphism between the stalk at $z_0$ and $\mathbb C$, given by $g\mapsto g(z_0)$. (They are isomorphic as fields... and I don't understand the "non-canonical" part, unless "canonical" refers to something that can be carried out globally?).