## Holomorphic Maps and Functions (Pt. III)

**Point of Post: **This is a continuation of this post.

We need a slightly weaker lemma before we prove the general identity theorem.

In all that follows we let, for functions , be the *agreement* set . Note that is closed, indeed since is Hausdorff we know that the diagonal is closed in . Now, by the definition of the product topology, the map given by is continuous (of course, are continuous since they are smooth maps and thus continuous!), and the closedness of follows since is just .

**Lemma: ***Let and be Riemann surfaces and a holomorphic maps. Then, if contains an open set, then .*

**Proof: **Let

Evidently is open, for if and is the guaranteed neighborhood of for which and agree, then by definition . Now, we claim that is also closed. Indeed, suppose that is a limit point of . Since and is a limit point of , we know that is a limit point of , and since is closed this implies that so that . We claim that there exists a coordinate disc at and a chart such that . Indeed, choose any chart at and any chart at and note that is an open neighborhood of . Find a coordinate disc at contained in and note then that evidently .

So, now, we know that since is a limit point of there exists an open set such that . Now, it’s easy to see then that is an open subset of such that . Since is connected, it follows from the Identity Theorem on that on and thus –thus . Since was arbitrary we see that contains all of its limit points, and is thus clopen. The coup de grace is then noticing that since, by assumption, contains an open set we see that is non-empty. Thus, by connectedness of we see that .

From this we can prove the Identity Theorem:

**Theorem(Identity Theorem): ***Let and be Riemann surfaces and holomorphic. If has a limit point, then .*

**Proof: **Let be a limit point of . As in the lemma we know that and that we can find a coordinate disc at and a chart at such that . We see then that is a disc and holomorphic functions on . Now, we claim that is a limit point of the agreement set . Indeed, if is any neighborhood of we know that is a neighborhood of , and thus contains a point , evidently is a point of . Since was arbitrary we see that is indeed a limit point of the agreement set of the coordinate representations of and . By the standard Identity Theorem we may conclude that , and thus we see that . The lemma now implies that as desired.

From this we get the following, extremely important, corollary:

**Corollary(Discreteness of Fibers): ***Let and be Riemann surfaces and be holomorphic and non-constant. Then, is discrete for any . In particular, if is compact, is finite for all .*

*Remark: *The Identity Theorem, and thus the discreteness of fibers, really only depended on being a Riemann surface, can be any one dimensional complex manifold.

We now prove the second most remarkable theorem in basic complex analysis (concerning the mapping properties of holomorphic functions, that is!)–the Open Mapping theorem.

**Theorem(Open Mapping Theorem): ***Let and be Riemann surfaces and holomorphic and non-constant. Then, is an open mapping.*

**Proof: **Let be open. For each choose such that . Choose then charts and around and such that and . We see then that is a holomorphic mapping . Now, by the Open Mapping theorem for mappings between domains we know that is open inside , and so is open in , and so open in . Thus, we see that is a neighborhood of in contained in . Since was arbitrary we see that is open as required.

From this we arrive at the theorem that should give an indication of how strong (strict, powerful, whatever word you’d like) the combination of compactness and holomorphicity is:

**Corollary: ***Let and be Riemann surfaces, with compact. If is holomorphic and non-constant, then is surjective and is compact.*

**Proof: **By the open mapping theorem we know that is open in , but it is also compact and so closed. From the connectedness of we get that and so is compact.

*Remark: *We remark again that connectivity of all spaces in sight in the above isn’t really necessary. The Open Mapping theorem (with the proof given) is true for holomorphic maps between one dimensional complex manifolds. The corollary also only required that be connected.

This allows us to give a quick proof of the fundamental theorem of algebra:

**Theorem(Fundamental Theorem of Algebra): ***Every non-constant polynomial over , has a root in .*

**Proof: **Let be a polynomial in . We know then that lifts to a (non-constant) holomorphic map by defining . Now, since is compact and connected we see that this implies that is surjective. In particular, there exists such that , and since we see that actually .

**References:**

[1] Varolin, Dror. *Riemann Surfaces by Way of Complex Analytic Geometry*. Providence, RI: American Mathematical Society, 2011. Print.

[2] Miranda, Rick. *Algebraic Curves and Riemann Surfaces*. Providence, RI: American Mathematical Society, 1995. Print.

[3] Forster, Otto. *Lectures on Riemann Surfaces*. New York: Springer-Verlag, 1981. Print.

[4] Conway, John B. *Functions of One Complex Variable*. New York: Springer-Verlag, 1978. Print.

[5] Gong, Sheng. *Concise Complex Analysis*. Singapore: World Scientific, 2001. Print

[6] Rudin, Walter. *Real and Complex Analysis.* New York: McGraw-Hill, 1966. Print.

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