Fitting a statement in a long list of equivalent results

Question

Theorem. Suppose that $D\subset \mathbb C$ is a connected open set. The following are equivalent.

1. Either $D=\mathbb C$ or $D$ is conformally equivalent to $\mathbb D$.
2. $D$ is homeomorphic to $\mathbb D$.
3. $D$ is simply connected.
4. Ind($\gamma,z)=0$ for every smooth closed curve $\gamma$ in $D$ and every $z\in\mathbb C\backslash D$.
5. $\mathbb C_{\infty}\backslash D$ is connected.
6. If $f\in H(D)$ then there exists a sequence of polynomials $(P_{n})$ such that $P_{n}\to f$ uniformly on compact subsets of $D$.
7. $\int_{\gamma} f(z)\,dz=0$ for every $f\in H(D)$ and every smooth closed curve $\gamma$ in $D$.
8. If $f\in H(D)$ then there exists $F\in H(D)$ with $F'=f$.
9. If $u:D\to \mathbb R$ is harmonic then $u=\mathbb R$e$(f)$ for some $f\in H(D)$.
10. If $f\in H(D)$ has no zero in $D$ then there exists $L\in H(D)$ with $f=e^{L}$.
11. If $f\in H(D)$ has no zero in $D$ then there exists $g\in H(D)$ with $g^{2}=f$.

The proof to this theorem can be found in David C. Ullrich's book Complex Made Simple, and it is surprisingly nice that it was proven as a chain (as in $1\implies 2\implies\cdots\implies 11\implies 1$).

Let $D$ and $G$ be connected open subsets of $\mathbb C$ with $D\subset G$. If $(f,D)$ is a function element which admits continuation along every path in $G$ which starts at a point of $D$ then we say that $(f,D)$ admits unrestricted continuation in $G$.

Now, consider the following:

Proposition. Let $D$ be an open connected subset of the plane. $D$ is simply connected if and only if for every functional element $(f,A)$, $A\subset D$ that admits unrestricted continuation in $D$ there exists a function $F\in H(D)$ such that $f=F\rvert_{A}$.

Between what numbers in the theorem would this proposition fit the best, in order to obtain a chain-like proof of the theorem?

Answer 1

I think a good place for the propostion would be between (7) and (8):

• Fix a base point $z_0\in D$. If $f$ admits unrestricted continuation in $D$, then that gives us a way to try to define $F\in H(D)$: draw a path from $z_0$ to $z\in D$ and define $F(z)$ according to the continuation of $f$ along that path. (7) allows one to prove that this definition is well defined (independent of the path).
• Again fix a base point $z_0$, and note that $\int_\gamma f(w)\,dw$ where $\gamma$ is any path from $z_0$ to $z$ gives unrestricted continuation in $D$ of the function element that equals $0$ at $z_0$. (This isn't quite right maybe, do we need an analytic function in a neighborhood of $z_0$...?) Then there is $F\in H(D)$ that extends this function, and one shows that $F'=f$.