How to prove that $D_2$ conformally equivalent to $D_1$ simply connected is simply connected If we couldn't use the topological properties of simply connectedness and conformal maps, how can we prove the following:

Let $D_1$ be a simply connected subset of $\Bbb C$, $D_2$ another subset of $\Bbb C$ such that there exists a conformal map $f:D_1\to D_2$. Then $D_2$ is simply connected.

I have an idea. We can consider $F_1$ the integral of $f$ in $D_1$ and $F_2$ the integral of $f$ in $D_2$. Then, by the uniqueness of the integrals, we know that there exists $c\in\Bbb C$ such that $F_2(z) = F_1(z) + c$ for all $z\in D_1\cap D_2$. We can define
\begin{equation*}
G(z) = \begin{cases}
F_1(z)+c & \quad \text{, if $z\in D_1$}\\
F_2(z) & \quad\text{, if $z\in D_2$}
\end{cases}
\end{equation*}
I should prove that $G(z) = 0$ for all $z$, knowing that $F_1(z) = 0$ for all $z$. Is it enough to see that $c = 0$ and we would have it? If so, how can we prove it?
Thanks in advance.

We consider $f$ a conformal map if it is holomorphic, injective in $D_1$ and $f(D_1) = D_2$. And we say that $D$ is simply connected if all cycle $\Gamma\sim 0 \;mod\; D$.


A result that I used in here was that if we consider $D$ a domain if $\Bbb C$, $D$ is simply connected iff for all $f$ holomorphic in $D$ and all cycle $\Gamma$ in $D$, $\int_{\Gamma}f(z)dz = 0$.

 A: A proof using the characterization

A domain $D \subset \Bbb C$ is simply connected if and only if $\int_\Gamma f(z) \, dz = 0$ for all cycles $\Gamma$ in $D$ and for all functions $f$ which are holomorphic in $D$.

works as follows:
Assume that $D_1$ and $D_2$ are domains in $\Bbb C$ such that $D_1$ is simply connected, and there is a conformal map $\phi$ from $D_1$ onto $D_2$. Note that the inverse function $\phi^{-1}$ is a conformal mapping from $D_2$ onto $D_1$.
Now let $f$ be holomorphic in $D_2$ and $\Gamma$ be a cycle in $D_2$. Then $\gamma =  \phi^{-1} \circ \Gamma$ is a cycle in $D_1$, and
$$ \tag{*}
 \int_\Gamma f(w) \, dw = \int_\gamma f(\phi(z)) \phi'(z) \, dz = 0
$$
because $f(\phi(z)) \phi'(z)$ is holomorphic in $D_1$ and $D_1$ is assumed to be simply connected.

Remark: If you are not convinced about the identity $(*)$ then choose a parametrization $\gamma:[0, 1] \to D_1$ of $\gamma$ and note that both sides are equal to
$$
 \int_0^1 f(\phi(\gamma(t))) \phi'(\gamma(t)) \gamma'(t) \, dt \, .
$$

With respect to your approach: I don't think that will work. You have to start with a holomorphic function $f$ and a cycle $\Gamma$ in $D_2$, and then you cannot talk of the “integral of $f$ in $D_1$.” Also $ D_1\cap D_2$ may be empty.
