# Abstract Nonsense

## Holomorphic Maps and Functions (Pt. III)

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

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

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In all that follows we let, for functions $f,g:X\to Y$, $A(f,g)$ be the agreement set $A(f,g)=\{x\in X:f(x)=g(x)\}$. Note that $A(f,g)$ is closed, indeed since $Y$ is Hausdorff we know that the diagonal $\Delta_Y=\{(y,y):y\in Y\}$ is closed in $Y\times Y$. Now, by the definition of the product topology, the map $h:X\to Y\times Y$ given by $x\mapsto (f(x),g(x))$ is continuous (of course, $f,g$ are continuous since they are smooth maps and thus continuous!), and the closedness of  $A(f,g)$ follows since $A(f,g)$ is just $h^{-1}(\Delta_Y)$.

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Lemma: Let $X$ and $Y$ be Riemann surfaces and $f,g:X\to Y$ a holomorphic maps. Then, if $A(f,g)$ contains an open set, then $A(f,g)=X$.

Proof: Let

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$S=\{x\in X:\text{ there exists a neighborhood }U\text{ of }x\text{ such that }f\mid_{U}=g\mid_{U}\}$

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Evidently $S$ is open, for if $x\in S$ and $U$ is the guaranteed neighborhood of $x$ for which $f$ and $g$ agree, then by definition $x\in U\subseteq S$. Now, we claim that $S$ is also closed. Indeed, suppose that $p\in X$ is a limit point of $S$. Since $S\subseteq A(f,g)$ and $p$ is a limit point of $S$, we know that $p$ is a limit point of $A(f,g)$, and since $A(f,g)$ is closed this implies that $p\in A(f,g)$ so that $f(p)=g(p)$. We claim that there exists a coordinate disc $(D,\varphi)$ at $p$ and a chart $(V,\psi)$ such that $f(D)\cap g(D)\subseteq V$. Indeed, choose any chart $(U,\tau)$ at $p$ and any chart $(V,\psi)$ at $f(p)=g(p)$ and note that $U\cap f^{-1}(V)\cap g^{-1}(V)$ is an open neighborhood of $p$. Find a coordinate disc $(D,\varphi)$ at $p$ contained in $U\cap f^{-1}(V)\cap g^{-1}(V)$ and note then that evidently $g(D)\cap f(D)\subseteq V$.

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So, now, we know that since $p$ is a limit point of $S$ there exists an open set $O\subseteq D$ such that $O\subseteq A(f,g)$. Now, it’s easy to see then that $\varphi(O)$ is an open subset of $\varphi(D)$ such that $\varphi(O)\subseteq A(\psi\circ f\circ\varphi^{-1},\psi\circ g\circ\varphi^{-1})$. Since $\varphi(D)$ is connected, it follows from the Identity Theorem on $\mathbb{C}$ that $\psi\circ f\circ\varphi^{-1}=\psi\circ g\circ\varphi^{-1}$ on $\varphi(D)$ and thus $p\in D\subseteq A(f,g)$–thus $p\in S$. Since $p$ was arbitrary we see that $S$ contains all of its limit points, and is thus clopen. The coup de grace is then noticing that since, by assumption, $A(f,g)$ contains an open set we see that $S$ is non-empty. Thus, by connectedness of $X$ we see that $A(f,g)\supseteq S=X$. $\blacksquare$

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From this we can prove the Identity Theorem:

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Theorem(Identity Theorem): Let $X$ and $Y$ be Riemann surfaces and $f,g:X\to Y$ holomorphic. If $A(f,g)$ has a limit point, then $A(f,g)=X$.

Proof: Let $p\in X$ be a limit point of $A(f,g)$. As in the lemma we know that $p\in A(f,g)$ and that we can find a coordinate disc $(D,\varphi)$ at $p$ and a chart $(U,\psi)$ at $f(p)=g(p)$ such that $f(D)\cap g(D)\subseteq V$. We see then that $\varphi(D)$ is a disc and $\psi\circ f\circ\varphi^{-1},\psi\circ g\circ\varphi^{-1}$ holomorphic functions on $\varphi(D)$. Now, we claim that $\varphi(p)$ is a limit point of the agreement set $A(\psi\circ f\circ\varphi^{-1},\psi\circ g\circ\varphi^{-1})$. Indeed, if $O$ is any neighborhood of $\varphi(p)$ we know that $\varphi^{-1}(O)$ is a neighborhood of $p$, and thus contains a point $x\in A(f,g)$, evidently $\varphi(x)$ is a point of $A(\psi\circ f\circ\varphi^{-1},\psi\circ g\circ\varphi^{-1})$. Since $O$ was arbitrary we see that $\varphi(p)$ is indeed a limit point of the agreement set of the coordinate representations of $f$ and $g$. By the standard Identity Theorem we may conclude that $A(\psi\circ f\circ\varphi^{-1},\psi\circ g\circ\varphi^{-1})=\varphi(D)$, and thus we see that $A(f,g)\supseteq D$. The lemma now implies that $A(f,g)=X$ as desired. $\blacksquare$

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

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Corollary(Discreteness of Fibers): Let $X$ and $Y$ be Riemann surfaces and $f:X\to Y$ be holomorphic and non-constant. Then, $f^{-1}(y)$ is discrete for any $y\in Y$. In particular, if $X$ is compact, $f^{-1}(y)$ is finite for all $y\in Y$.

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

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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.

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Theorem(Open Mapping Theorem): Let $X$ and $Y$ be Riemann surfaces and $f:X\to Y$ holomorphic and non-constant. Then, $f$ is an open mapping.

Proof: Let $W\subseteq X$ be open. For each $y\in f(W)$ choose $x\in W$ such that $f(x)=y$. Choose then charts $(U,\varphi)$ and $(V,\psi)$ around $x$ and $y$ such that $U\subseteq W$ and $f(U)\subseteq V$. We see then that $\psi\circ f\circ\varphi^{-1}$ is a holomorphic mapping $\varphi(U)\to \psi(V)$. Now, by the Open Mapping theorem for mappings between domains we know that $(\psi\circ f\circ\varphi^{-1})(\varphi(U))=\psi(f(U))$ is open inside $\psi(V)$, and so $f(U)$ is open in $V$, and so open in $Y$. Thus, we see that $f(U)$ is a neighborhood of $y$ in $Y$ contained in $f(W)$. Since $y$ was arbitrary we see that $f(W)$ is open as required. $\blacksquare$

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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:

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Corollary: Let $X$ and $Y$ be Riemann surfaces, with $X$ compact. If $f:X\to Y$ is holomorphic and non-constant, then $f$ is surjective and $Y$ is compact.

Proof: By the open mapping theorem we know that $f(X)$ is open in $Y$, but it is also compact and so closed. From the connectedness of $Y$ we get that $f(X)=Y$ and so $Y$ is compact. $\blacksquare$

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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 $Y$ be connected.

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

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Theorem(Fundamental Theorem of Algebra): Every non-constant polynomial over $\mathbb{C}$, has a root in $\mathbb{C}$.

Proof: Let $f(x)$ be a polynomial in $\mathbb{C}[x]$. We know then that $f(x)$ lifts to a (non-constant) holomorphic map $\mathbb{C}_\infty\to\mathbb{C}_\infty$ by defining $\displaystyle f(\infty)=\lim_{z\to\infty}f(z)=\infty$. Now, since $\mathbb{C}_\infty$ is compact and connected we see that this implies that $f$ is surjective. In particular, there exists $z_0\in\mathbb{C}_\infty$ such that $f(z_0)=\infty$, and since $f(\infty)$ we see that actually $z_0\in\mathbb{C}$. $\blacksquare$

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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.