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The monodromy theorem gives equality of continuations that exist under its hypothesis. It does not assert that every holomorphic function admits analytic continuation everywhere, even within a simply connected domain.

For an explicit obstruction, take $f(z)=1/(1-z)$ near $a=0$ and attempt to apply the theorem on $D=\mathbb{C}$. The path $\eta (t)=t$ ends at $1$, where no terminal germ of a holomorphic function exists. If a holomorphic representative $G$ near $1$ continued $f$, then the identity theorem would extend $(1-z)G(z)=1$ to that neighbourhood. Substituting $z=1$ would give $0=1$. Thus the pole prevents the existence hypothesis, although $\mathbb{C}$ is a simply connected domain. The analytic continuation article explains why the same rational function does continue throughout $\mathbb{C}\setminus \{1\}$.

Existence at an endpoint by some route is also weaker than existence along every path from the base point. Along the real segment from $1$ towards $0$, a holomorphic square root of $z$ tends to $0$, yet cannot continue holomorphically through $0$: a local representative $G$ would satisfy $G^2=z$ by the identity theorem, and differentiating at $0$ would give $2G(0)G^{\prime }(0)=1$ with $G(0)=0$. A finite limiting value therefore does not remove a branch point. More severe obstructions appear at a natural boundary of a holomorphic function.

Along one path, analytic continuation is a chain of function elements on overlapping open discs, each agreeing with the preceding germ of a holomorphic function near the crossing point. The identity theorem makes this chain unique along that fixed path. Equality at a crossing means equality on a neighbourhood, not just equality of a single value.

For the deformation step, fix one intermediate path $t\mapsto H(s_0,t)$ and its continuation. By compactness of $[0,1]$, choose a finite chain of representative open discs and subdivide the parameter so that each subarc fits in its assigned open disc, with consecutive junctions in overlap neighbourhoods where the representatives agree. Joint continuity of $H$ ensures that every sufficiently nearby intermediate path uses that same chain and those same overlaps. The endpoints stay fixed, so the terminal germ of a holomorphic function stays unchanged. Existence along every intermediate path supplies this argument at each $s_0$; local constancy and connectedness of $[0,1]$ give equality at $s=0$ and $s=1$. This proves the monodromy theorem without assuming that every analytic germ projection is a covering map.

Start at $a=1$ with complex logarithm germ of a holomorphic function $L(1)=0$ and holomorphic square root germ of a holomorphic function $S(1)=1$. On the punctured plane $D=\mathbb{C}\setminus \{0\}$, both admit analytic continuation along every path. To construct it locally, for any $c\ne 0$ choose a small open disc avoiding $0$, select a holomorphic logarithm there, and select its additive multiple of $2\pi i$ to agree with the arriving germ of a holomorphic function; exponentiating half this logarithm supplies the arriving holomorphic square root. A finite chain of these open discs follows any given path.

Compare two paths from $1$ to $i$, parametrized by $0\le t\le 1$:The first follows a counterclockwise quarter-circle; the second follows a clockwise three-quarter-circle. Following ${\gamma }_A$ and then path reversal of ${\gamma }_B$ makes one positive turn about $0$. Its winding number is $1$, so the two paths cannot be homotopic with endpoints fixed in the punctured plane.

The continued values along the chosen paths areAt their common endpoint,Both complex logarithm values exponentiate to $i$, and both holomorphic square root values square to $i$. These give different terminal germ of a holomorphic function representatives, consistently obtained from the same starting germ of a holomorphic function. The monodromy theorem applies, but does not equate these two non-homotopic paths.

The closed path $\lambda (t)=e^{2\pi it}$ instead returns to $1$, with continued values $L_{\lambda }(t)=2\pi it$ and $S_{\lambda }(t)=e^{\pi it}$. One counterclockwise turn changes the complex logarithm by $2\pi i$ and flips the holomorphic square root from $1$ to $-1$. Hence neither initial germ of a holomorphic function can come from a global holomorphic logarithm or a global holomorphic square root of $z$ on the punctured plane. Repeating a loop of winding number $n$ gives the changes $2\pi in$ and $(-1{)}^n$. ^[ref]

Now restrict to the slit domain $D_{\mathrm{slit}}=\mathbb{C}\setminus (-\infty ,0]$. The principal logarithm and $\exp (\mathrm{Log}z/2)$ realize the same initial germ of a holomorphic function for each expression globally. To see that this is a simply connected domain, polar form identifies this domain with $(0,\infty )\times (-\pi ,\pi )$, a convex set; every loop contracts there. The simply connected corollary of the monodromy theorem now forces equality along every pair of paths with matching endpoints. In particular, ${\gamma }_A$ and the segment $\sigma (t)=1-t+it$ both give $L(i)=\pi i/2$ and $S(i)=(1+i)/\sqrt{2}$. The explicit homotopy $H(s,t)=(1-s){\gamma }_A(t)+s\sigma (t)$ stays in the closed first quadrant away from $0$, hence in $D_{\mathrm{slit}}$.

The other route ${\gamma }_B$ crosses the removed ray at $t=2/3$ and is no longer admissible. In the punctured plane, that crossing is locally regular: analytic continuation passes through $-1$ smoothly, although evaluating the principal logarithm would reset its argument. The branch cut changes the allowed domain; it does not create additional branch points or poles. That distinction is the beauty of the obstruction: perfectly compatible local holomorphic functions can fail to make a global choice solely because of the allowed based loops. The branches of a multivalued function article develops how moving the branch cut changes the selected values.

The monodromy theorem says that, when analytic continuation exists throughout a domain, its ambiguity depends on homotopy. Let $D\subset \mathbb{C}$ be a domain, $a\in D$, and $g_a$ a germ of a holomorphic function at $a$. Assume $g_a$ admits analytic continuation along every continuous path in $D$ starting at $a$. Two such paths ending at $b$ give the same terminal germ of a holomorphic function if they are homotopic with endpoints fixed in $D$. ^[ref]

Here a homotopy with endpoints fixed is a continuous map $H:[0,1{]}^2\to D$ satisfyingEvery intermediate path must stay in $D$. Equality means agreement on a neighbourhood of $b$, rather than merely equal values at $b$. ^[ref]

The identity theorem provides local rigidity; the monodromy theorem shows precisely how topology controls the remaining global freedom. Deforming a route without crossing a forbidden point cannot change the resulting germ of a holomorphic function. Thus a complicated analytic continuation calculation can be replaced by one on a simpler homotopic route. Distinct homotopy classes can still give the same continuation: a rational function on its domain already gives a single-valued answer along every route, even around a pole.

The monodromy theorem states that a germ admitting analytic continuation along every path in a simply connected domain extends to a single-valued holomorphic function there. Equivalently continuations along homotopic paths with fixed endpoints agree. For a biholomorphism between the two tori, lift its composition with $\mathbb{C}\to \mathbb{C}/{\Lambda }_1$ through the second covering. Simple connectivity, or monodromy applied to local inverse charts, yields a holomorphic lift $F:\mathbb{C}\to \mathbb{C}$.

Lift the inverse similarly. The compositions of the lifts differ from the identity by lattice translations; adjusting the lifts shows that $F$ is bijective. By the allowed classification, $F(z)=az+b$ with $a\ne 0$. For $\lambda \in {\Lambda }_1$, its two arguments $z,z+\lambda$ represent the same torus point, so $a\lambda =F(z+\lambda )-F(z)\in {\Lambda }_2$. Thus $a{\Lambda }_1\subseteq {\Lambda }_2$. Apply the inverse lift to obtain the reverse inclusion. ThereforeThe translation $b$ affects the chosen origin, not the period lattice.

The monodromy theorem states that if a holomorphic germ can be analytically continued along every path in a domain, its continuations along paths homotopic with fixed endpoints coincide. In a simply connected domain this gives a single-valued holomorphic function on the whole domain.

Write ${\zeta }_j=e^{2\pi ij/n}$. Remove the radial rays $R_j=\{t{\zeta }_j:t\ge 1\}$ and set $U=\mathbb{C}\setminus {\bigcup }_j{R}_j$. This is an open star-shaped domain containing zero, hence simply connected, and $h=z^n-1$ never vanishes there. Locally its logarithm continues along all paths in $U$, so monodromy gives a holomorphic logarithm $L$ andEquivalently, integrate $h^{\prime }/h$ on the simply connected domain and choose the additive constant to obtain $e^L=h$.

Take $r$ copies of the slit domain, labeled by $j\in \mathbb{Z}/r\mathbb{Z}$, carrying values $e^{(L+2\pi ij)/r}$. Across each cut identify banks so that a positive loop around its simple zero increments the sheet index by one. This cyclic gluing produces the unbranched Riemann surface over $\mathbb{C}\setminus Z$, on which $w^r=z^n-1$ is single valued. Completing each finite puncture gives local coordinate $w$, withThere is one ramification point of index $r$ over each of the $n$ roots of unity.

At infinity a loop around all $n$ zeros acts on sheet labels by $j\mapsto j+n(\mathrm{mod}r)$. With $d=\gcd (n,r)$ this permutation has $d$ cycles, each of length $r/d$. In the compactification there are therefore $d$ points over infinity, each with ramification index $r/d$. Infinity is unbranched precisely when $d=r$, namely when $r\mid n$. Counting branch values in the base sphere,If “branch points” denotes ramification points upstairs instead, the count is $n$ or $n+d$ respectively, with exactly the same stated equivalence.

In local coordinates centred at $p$ and $f(p)$, a nonconstant rational map has the formThe point $p$ is a ramification point of a holomorphic map when its ramification index of a holomorphic map satisfies $e_p>1$. A branch value of a holomorphic map, also called a branch point in the target, is a value $f(p)$ of some ramification point.

Let $B$ be the finite set of branch values and put $R=f^{-1}(B)$. For every $w\notin B$, all points of $f^{-1}(w)$ have local degree one, so the holomorphic inverse function theorem supplies disjoint neighbourhoods on which $f$ is biholomorphic. Compactness of the fibre lets their target neighbourhoods be intersected to one evenly covered neighbourhood of $w$. Henceis an unramified covering map. Here the displayed restriction requires $R=f^{-1}(B)$; if “ramification point” is reserved only for points with $e_p>1$, then the source deletion must be written $f^{-1}(B)$.

The monodromy theorem says that analytic continuations along endpoint-fixed homotopic paths have the same terminal germ. Equivalently here, the path lifting theorem lifts a loop $\gamma$ based at $w\notin B$ from each point of $f^{-1}(w)$; taking the endpoint of each lift permutes that fibre. The permutation depends only on $[\gamma ]\in {\pi }_1({\mathbb{C}}_{\infty }\setminus B,w)$, giving the monodromy group of a covering.

For the stated function, make the Möbius change of coordinateThenThe critical points are $t=0,\pm 1/\sqrt{2},\infty$, with branch valuesA loop around $0$ interchanges the two roots born from $t=0$, so its branch cycle is a transposition. A loop around $-1/4$ simultaneously interchanges the two pairs that collide there, so its branch cycle is a product of two disjoint transpositions. With a suitable labelling these areThey generate a transitive group of order eight; their product is a four-cycle. Therefore the full monodromy group isthe dihedral group in its action on the four vertices of a square.

An analytic continuation along $\alpha$ is a chain of function elements $(f_t,U_t)$ covering the path, beginning with the germ of $(f,U)$ at $\alpha (0)$, such that neighboring elements agree on the component of overlap met by the path. The terminal element is $(g,V)$.

The monodromy theorem says that if continuation is possible along every path in $D$, then continuations along endpoint-fixed homotopic paths have the same terminal germ. It follows by subdividing a homotopy square into small rectangles on which the identity theorem identifies neighboring germs. Hence

The requested final example with homotopic paths and different terminal values cannot exist under these hypotheses; it contradicts the theorem just proved. With “non-homotopic” in place of “homotopic”, take the germ of $\log z$ near $1$ in ${\mathbb{C}}^{\ast }$. Continuation along the constant path gives $0$, whereas continuation once counterclockwise around zero gives $2\pi i$.

On a simply connected domain $D$, a germ of a holomorphic function admitting analytic continuation along every path from its base point determines a unique holomorphic function $F:D\to \mathbb{C}$ with that initial germ of a holomorphic function. ^[ref]

Indeed, any two paths with the same endpoints are homotopic with endpoints fixed. Define $F(b)$ using the terminal germ of a holomorphic function. The monodromy theorem makes this independent of the path. In a small open disc about $b$, extend the chosen route by a short segment; the same local representative supplies $F$ throughout the open disc, proving holomorphy. The identity theorem proves uniqueness. The existence hypothesis of the monodromy theorem must still be checked.