[proofplan] We show that analytic continuation along any two paths from $a$ to a point $b \in \Omega$ yields the same function element at $b$. The key input is that $\Omega$ is simply connected, so any two paths from $a$ to $b$ are homotopic with fixed endpoints. We prove that analytic continuation is invariant under homotopy: if a family of paths continuously deforms one into the other, the resulting function elements at the endpoint agree. This gives a well-defined holomorphic function on $\Omega$ by declaring its value at each point to be the result of continuation along any path from $a$. [/proofplan]

[step:Establish that analytic continuation along homotopic paths yields the same germ]Let $\gamma_0, \gamma_1: [0,1] \to \Omega$ be two paths from $a$ to $b$. Since $\Omega$ is simply connected, there exists a homotopy $H: [0,1] \times [0,1] \to \Omega$ with $H(t,0) = \gamma_0(t)$, $H(t,1) = \gamma_1(t)$, $H(0,s) = a$, and $H(1,s) = b$ for all $s, t \in [0,1]$. For each $s \in [0,1]$, define the path $\gamma_s(t) := H(t,s)$. By hypothesis, $f$ can be analytically continued along each $\gamma_s$. Let $(f_s, B_s)$ denote the function element obtained at $b = \gamma_s(1)$ by continuation along $\gamma_s$, where $f_s$ is holomorphic on a disc $B_s$ centred at $b$. [claim:The germ of $f_s$ at $b$ is locally constant in $s$] For each $s_0 \in [0,1]$, there exists $\delta > 0$ such that for all $s$ with $|s - s_0| < \delta$, the function elements $f_s$ and $f_{s_0}$ agree on a neighbourhood of $b$. [/claim] [proof] Fix $s_0 \in [0,1]$. The analytic continuation of $f$ along $\gamma_{s_0}$ is given by a chain of function elements $(f_0^{(0)}, D_0), (f_1^{(0)}, D_1), \ldots, (f_N^{(0)}, D_N)$ covering $\gamma_{s_0}$, where each $D_k$ is an open disc, $\gamma_{s_0}([t_k, t_{k+1}]) \subset D_k$, and $f_k^{(0)}$ agrees with $f_{k+1}^{(0)}$ on $D_k \cap D_{k+1}$. Since the image $H([0,1] \times [0,1])$ is compact in $\Omega$ and each $D_k$ is open, there exists a Lebesgue number $\lambda > 0$ for the cover $\{D_k\}$ relative to the paths near $\gamma_{s_0}$. By continuity of $H$, there exists $\delta > 0$ such that for $|s - s_0| < \delta$, the path $\gamma_s$ remains within the union $D_0 \cup D_1 \cup \cdots \cup D_N$, with $\gamma_s([t_k, t_{k+1}])$ still contained in $D_k$ for each $k$. Under this condition, the same chain of function elements $(f_0^{(0)}, D_0), \ldots, (f_N^{(0)}, D_N)$ provides an analytic continuation of $f$ along $\gamma_s$. By the uniqueness of analytic continuation along a given path (once the chain of discs is fixed), the function element at $b$ is the same as $(f_N^{(0)}, D_N)$. Therefore $f_s = f_{s_0}$ on $D_N$, and in particular their germs at $b$ agree. [/proof][/step]

[guided]The goal of this step is to show that the germ at $b$ obtained by analytic continuation does not change when we continuously deform the path. The key idea is that a small deformation of a path cannot "jump" the analytic continuation to a different branch, because the identity theorem forces holomorphic function elements that agree on an open set to be identical everywhere on their common domain. Let $\gamma_0, \gamma_1: [0,1] \to \Omega$ be two paths from $a$ to $b$. Since $\Omega$ is simply connected, there exists a homotopy $H: [0,1] \times [0,1] \to \Omega$ with $H(t,0) = \gamma_0(t)$, $H(t,1) = \gamma_1(t)$, $H(0,s) = a$, and $H(1,s) = b$ for all $s, t \in [0,1]$. For each $s \in [0,1]$, define the intermediate path $\gamma_s(t) := H(t,s)$. By hypothesis, $f$ can be analytically continued along each $\gamma_s$. Let $(f_s, B_s)$ denote the function element obtained at $b = \gamma_s(1)$ by continuation along $\gamma_s$, where $f_s$ is holomorphic on a disc $B_s$ centred at $b$. We now prove the crucial claim: the germ of $f_s$ at $b$ is locally constant in $s$. Fix $s_0 \in [0,1]$. The analytic continuation of $f$ along $\gamma_{s_0}$ is given by a chain of function elements $(f_0^{(0)}, D_0), (f_1^{(0)}, D_1), \ldots, (f_N^{(0)}, D_N)$ covering $\gamma_{s_0}$. Here each $D_k$ is an open disc, the path satisfies $\gamma_{s_0}([t_k, t_{k+1}]) \subset D_k$ for a partition $0 = t_0 < t_1 < \cdots < t_N < t_{N+1} = 1$, and the compatibility condition holds: $f_k^{(0)}$ agrees with $f_{k+1}^{(0)}$ on the overlap $D_k \cap D_{k+1}$ for each $k = 0, \ldots, N-1$. Why can nearby paths use the same chain? Since the image $H([0,1] \times [0,1])$ is compact in $\Omega$ and the discs $\{D_0, D_1, \ldots, D_N\}$ form an open cover of the compact set $\gamma_{s_0}([0,1])$, there exists a Lebesgue number $\lambda > 0$ for this cover. By continuity of $H$, there exists $\delta > 0$ such that for all $s$ with $|s - s_0| < \delta$, the uniform distance satisfies $\sup_{t \in [0,1]} |H(t,s) - H(t,s_0)| < \lambda$. This ensures that $\gamma_s([t_k, t_{k+1}])$ remains contained in $D_k$ for each $k = 0, \ldots, N$. Under this condition, the same chain of function elements $(f_0^{(0)}, D_0), (f_1^{(0)}, D_1), \ldots, (f_N^{(0)}, D_N)$ provides a valid analytic continuation of $f$ along $\gamma_s$. Why? At the starting disc $D_0$, the function element $f_0^{(0)}$ is the original germ of $f$ at $a$, which is the same regardless of which path we follow. At each subsequent disc $D_k$, the function element $f_k^{(0)}$ is the unique holomorphic function on $D_k$ that agrees with $f_{k-1}^{(0)}$ on the overlap $D_{k-1} \cap D_k$. This uniqueness follows from the identity theorem: two holomorphic functions on a connected open set (the disc $D_k$) that agree on a non-empty open subset ($D_{k-1} \cap D_k \neq \varnothing$) must be identical on all of $D_k$. By the uniqueness of analytic continuation along a given path once the chain of discs is fixed, the function element obtained at $b$ via continuation along $\gamma_s$ is $(f_N^{(0)}, D_N)$ — the same terminal element as for $\gamma_{s_0}$. Therefore $f_s = f_{s_0}$ on $D_N$, and in particular the germs of $f_s$ and $f_{s_0}$ at $b$ agree. This establishes that $s \mapsto [\text{germ of } f_s \text{ at } b]$ is locally constant on $[0,1]$. The key hypotheses used are: (1) the openness of each $D_k$, which provides room for the path to move without leaving the disc; (2) the identity theorem, which forces the function elements to match on each disc; and (3) the continuity of the homotopy $H$, which guarantees that small changes in $s$ produce only small changes in the path.[/guided]

[step:Conclude that the germ at $b$ is independent of the path] The map $s \mapsto [\text{germ of } f_s \text{ at } b]$ is locally constant on $[0,1]$ by the claim above. Since $[0,1]$ is connected, a locally constant function on $[0,1]$ is constant. Therefore the germ of $f_0$ at $b$ equals the germ of $f_1$ at $b$, which means the analytic continuation along $\gamma_0$ and along $\gamma_1$ produce the same function element at $b$. [/step]

[step:Define the global holomorphic function on $\Omega$]For each $b \in \Omega$, choose any path $\gamma: [0,1] \to \Omega$ with $\gamma(0) = a$ and $\gamma(1) = b$ (such a path exists since $\Omega$ is open and connected, hence path-connected). Define $\tilde{f}(b)$ to be the value at $b$ of the function element obtained by analytic continuation of $f$ along $\gamma$. By the preceding step, $\tilde{f}(b)$ is independent of the choice of $\gamma$, so $\tilde{f}: \Omega \to \mathbb{C}$ is well-defined. Moreover, at each point $b$, analytic continuation provides a holomorphic function element on an open disc around $b$ that agrees with $\tilde{f}$. Therefore $\tilde{f}$ is holomorphic on $\Omega$. Finally, in a neighbourhood of $a$, the continuation along the constant path produces the original function element $f$, so $\tilde{f}$ agrees with $f$ near $a$. This completes the proof.[/step]

[guided]We have established two ingredients: (1) analytic continuation along any path from $a$ to $b$ produces a well-defined germ at $b$ (independent of the path chosen), and (2) this continuation exists for every $b \in \Omega$ (by hypothesis). To see that $\tilde{f}$ is holomorphic: fix $b \in \Omega$ and let $\gamma$ be a path from $a$ to $b$. The continuation along $\gamma$ yields a function element $(g, D)$ with $g$ holomorphic on an open disc $D$ centred at $b$. For any $b' \in D$, we can extend $\gamma$ by a straight-line segment from $b$ to $b'$ (remaining inside $D$), and the continuation along this extended path gives $g(b')$. By path-independence, $\tilde{f}(b') = g(b')$. So $\tilde{f}$ agrees with the holomorphic function $g$ on the disc $D$, proving $\tilde{f}$ is holomorphic at $b$. Note where simple connectivity is essential: without it, two paths from $a$ to $b$ might not be homotopic, and the analytic continuation could genuinely produce different values (as happens with $\log z$ on $\mathbb{C} \setminus \{0\}$, which is not simply connected). Simple connectivity guarantees every loop is contractible, which forces path-independence of the continuation.[/guided]