Mixing
Entire one-to-one functions are linear
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Can we prove that every entire one-to-one function is linear?
complex-analysis
edited Jan 31 at 21:21
Did
249k23228466
asked Mar 29 '11 at 16:59
PeteyPetey
211143
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|
show 3 more comments
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Can we prove that every entire one-to-one function is linear?
complex-analysis
edited Jan 31 at 21:21
Did
249k23228466
asked Mar 29 '11 at 16:59
PeteyPetey
211143
$endgroup$
1
$begingroup$
en.wikipedia.org/wiki/Picard_theorem (which is overkill if this is homework or something)
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– yoyo
Mar 29 '11 at 17:06
2
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Maybe you can use the Weierstrass Factorization Theorem?
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– Adrián Barquero
Mar 29 '11 at 17:06
4
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By 1-1, do you mean injective or bijective?
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– lhf
Mar 29 '11 at 17:52
1
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@Arturo: you should let OPs do some of the fixing! It is a very instructive activity :)
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– Mariano Suárez-Álvarez
Mar 29 '11 at 19:56
1
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@Mariano: But it bugs me so...
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– Arturo Magidin
Mar 29 '11 at 20:40
|
show 3 more comments
$begingroup$
Can we prove that every entire one-to-one function is linear?
complex-analysis
edited Jan 31 at 21:21
Did
249k23228466
asked Mar 29 '11 at 16:59
PeteyPetey
211143
$endgroup$
Can we prove that every entire one-to-one function is linear?
complex-analysis
complex-analysis
edited Jan 31 at 21:21
Did
249k23228466
asked Mar 29 '11 at 16:59
PeteyPetey
211143
edited Jan 31 at 21:21
Did
249k23228466
asked Mar 29 '11 at 16:59
PeteyPetey
211143
edited Jan 31 at 21:21
Did
249k23228466
edited Jan 31 at 21:21
Did
249k23228466
edited Jan 31 at 21:21
Did
249k23228466
249k23228466
asked Mar 29 '11 at 16:59
PeteyPetey
211143
asked Mar 29 '11 at 16:59
PeteyPetey
211143
asked Mar 29 '11 at 16:59
PeteyPetey
211143
211143
1
$begingroup$
en.wikipedia.org/wiki/Picard_theorem (which is overkill if this is homework or something)
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– yoyo
Mar 29 '11 at 17:06
2
$begingroup$
Maybe you can use the Weierstrass Factorization Theorem?
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– Adrián Barquero
Mar 29 '11 at 17:06
4
$begingroup$
By 1-1, do you mean injective or bijective?
$endgroup$
– lhf
Mar 29 '11 at 17:52
1
$begingroup$
@Arturo: you should let OPs do some of the fixing! It is a very instructive activity :)
$endgroup$
– Mariano Suárez-Álvarez
Mar 29 '11 at 19:56
1
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@Mariano: But it bugs me so...
$endgroup$
– Arturo Magidin
Mar 29 '11 at 20:40
|
show 3 more comments
1
$begingroup$
en.wikipedia.org/wiki/Picard_theorem (which is overkill if this is homework or something)
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– yoyo
Mar 29 '11 at 17:06
2
$begingroup$
Maybe you can use the Weierstrass Factorization Theorem?
$endgroup$
– Adrián Barquero
Mar 29 '11 at 17:06
4
$begingroup$
By 1-1, do you mean injective or bijective?
$endgroup$
– lhf
Mar 29 '11 at 17:52
1
$begingroup$
@Arturo: you should let OPs do some of the fixing! It is a very instructive activity :)
$endgroup$
– Mariano Suárez-Álvarez
Mar 29 '11 at 19:56
1
$begingroup$
@Mariano: But it bugs me so...
$endgroup$
– Arturo Magidin
Mar 29 '11 at 20:40
1
1
$begingroup$
en.wikipedia.org/wiki/Picard_theorem (which is overkill if this is homework or something)
$endgroup$
– yoyo
Mar 29 '11 at 17:06
$begingroup$
en.wikipedia.org/wiki/Picard_theorem (which is overkill if this is homework or something)
$endgroup$
– yoyo
Mar 29 '11 at 17:06
2
2
$begingroup$
Maybe you can use the Weierstrass Factorization Theorem?
$endgroup$
– Adrián Barquero
Mar 29 '11 at 17:06
$begingroup$
Maybe you can use the Weierstrass Factorization Theorem?
$endgroup$
– Adrián Barquero
Mar 29 '11 at 17:06
4
4
$begingroup$
By 1-1, do you mean injective or bijective?
$endgroup$
– lhf
Mar 29 '11 at 17:52
$begingroup$
By 1-1, do you mean injective or bijective?
$endgroup$
– lhf
Mar 29 '11 at 17:52
1
1
$begingroup$
@Arturo: you should let OPs do some of the fixing! It is a very instructive activity :)
$endgroup$
– Mariano Suárez-Álvarez
Mar 29 '11 at 19:56
$begingroup$
@Arturo: you should let OPs do some of the fixing! It is a very instructive activity :)
$endgroup$
– Mariano Suárez-Álvarez
Mar 29 '11 at 19:56
1
1
$begingroup$
@Mariano: But it bugs me so...
$endgroup$
– Arturo Magidin
Mar 29 '11 at 20:40
$begingroup$
@Mariano: But it bugs me so...
$endgroup$
– Arturo Magidin
Mar 29 '11 at 20:40
|
show 3 more comments
6 Answers
6
active
oldest
votes
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You can rule out polynomials of degree greater than $1$, because the derivative of such a polynomial will have a zero by the fundamental theorem of algebra, and a holomorphic function is $(n+1)$-to-$1$ near a zero of its derivative of order $n$.
To finish, you need to rule out entire functions that are not polynomials. If $f$ is such a function, then $f(1/z)$ has an essential singularity at $z=0$. To see that this implies that $f$ is not one-to-one, you could apply Picard's theorem as yoyo indicates. Or you could proceed as follows. By Casorati-Weierstrass, $f({z:|z|>n})$ is dense in $mathbb{C}$ for each positive integer $n$. By the open mapping theorem, the set is open. By Baire's theorem, $D=bigcap_n f({z:|z|>n})$ is dense in $mathbb{C}$. In particular, $D$ is not empty, and every element of $D$ has infinitely many preimage points under $f$.
I just realized that there is an easier way to apply Casorati-Weierstrass, with no need for Baire. If $f$ is entire and not a polynomial, then $f({z:|z|<1})$ is open, and $f({z:|z|>1})$ is dense. Therefore these sets have nonempty intersection. Every element of the intersection has at least $2$ preimage points.
edited May 27 '13 at 4:54
Michael Hardy
1
answered Mar 29 '11 at 17:18
Jonas MeyerJonas Meyer
41.1k6148258
$endgroup$
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Nice application of Baire. Thanks!
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– Digital Gal
Mar 29 '11 at 19:24
add a comment |
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By shifting $z$, without loss of generality you can assume $f(0) = 0$. By the open mapping theorem, $f(z)$ maps some open set $U$ containing $0$ to another one, call it $V$. Since $f(z)$ is to be one-to-one, $f(z)$ can't map any $z$ outside of $U$ to $V$. Thus ${1 over f(z)}$ is bounded outside of $U$. Therefore ${z over f(z)}$ is an entire function that grows
no faster than linearly: $|{z over f(z)}| < A|z| + B$ for some $A$ and $B$.
It's easy from here to show that $g(z) = {z over f(z)}$ is linear; for any $z_0$
${g(z) - g(z_0) over z - z_0}$ must be bounded and therefore is a constant by Liouville's theorem. So ${z over f(z)} = c_1z + c_2$ for some $c_1$ and $c_2$. Hence $f(z) = {zover c_1z + c_2}$. Since $f(z)$ has no poles and is nonconstant, $c_1$ must be zero and $c_2$ nonzero. We conclude that $f(z) = {1 over c_2} z$.
answered Mar 29 '11 at 22:16
ZarraxZarrax
35.7k250104
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Why f(z) has simple zero at z=0?
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– nicksohn
Dec 30 '15 at 10:33
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Nice proof. In the beginning, you say wlog wma $f(z)=0$. Maybe you mean $f(0)=0$?
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– perlman
Oct 12 '17 at 5:55
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@nicksohn That follows from $f(z)$ being one to one
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– Zarrax
Oct 12 '17 at 17:40
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@MathewJames yes, thanks, corrected it.
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– Zarrax
Oct 12 '17 at 17:40
add a comment |
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Let $f:mathbb Ctomathbb C$ entire and injective. Let $U=f(mathbb C)$. $U$ is an open subset of the plane.
$U$ is simply connected: indeed, to check this it is enough to show that the integral of every analytic function on $U$ along every closed curved in $U$ is zero, and you can do this by "changing variables using $f$".
Next, if $Usubsetneqmathbb C$, from Riemann's theorem we know that there is an biholomorphic map $Uto D$, with $D$ the unit disc. Composing with $f$, we get a biholomorphic map $mathbb Cto D$, and this is impossible. We see then that $f$ is in fact bijective and, in fact, an homeomorphism. Composing with a translation, we can assume that $f(0)=0$.
Using this, one can see that the function $1/f(z)$ is bounded at $infty$ and has a simple pole at $0$, so $g(z)=z/f(z)$ is entire and bounded by a function of the form $cz$ for some constant $c$. Using Cauchy's estimates for the Taylor coefficients of $g$, we see that $g$ is a polynomial of very low degree. Translating this to information about $f$, we can conclude what we want.
(This avoids Picard but uses Riemann... :( )
answered Mar 29 '11 at 17:28
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
112k7159291
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How do we know that the pole at 0 is of order 1, from the fact that f(0)=0? What if 0 was zero of order n>1 for f(z)...thanks, @Mariano Suárez-Alvarez.
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– User001
Dec 18 '14 at 6:18
add a comment |
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I'll give the "usual" proof.
Note that by Little Picard, $f$ misses at most one point; but it is a homeomorphism onto its image, and the plane minus a point is not simply connected. Thus $f$ is onto $mathbb{C}$, and hence bijective. Then $f$ has a holomorphic inverse, which is enough to imply $f$ is proper, that is, the pre-image of a compact set is compact. This in turn implies
$$ lim_{zrightarrowinfty} f(z)=infty,$$
and thus if we define $f(infty)=infty$, $f$ becomes a Möbius transformation of the Riemann sphere. So $f$ has the form
$f(z) = frac{az+b}{cz+d},$
and it is easy to see that if $f$ is entire on $mathbb{C}$, then $c=0$.
edited Sep 29 '13 at 9:47
azimut
16.6k1052101
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I should also add that I believe there is a proof using just Little Picard and the Cauchy-Riemann equations. I just forget all the details :-). Basically you show $f$ is a Euclidean similarity on $mathbb{R}^2$, so it has the form $kAz+b$, where $A$ is $2times2$ orthogonal. Since $f$ is holomorphic, $Ain SO(2)$. Using the well known form of such matrices, you then show $A$ is rotation by some angle $theta$, so $f(z)=ke^{itheta}z+b$.
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– user641
Mar 29 '11 at 20:36
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Why is it that $c=0$ if $f$ is entire?
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– user225477
Feb 16 '17 at 18:33
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@Zermelo's_Choice, I think otherwise $f$ would have a pole at the point $z$ such that $cz+d=0$.
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– user135520
Apr 27 '18 at 16:43
add a comment |
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This proof uses only the Open Mapping Theorem and Cauchy's Estimates.
Suppose $f$ is entire and injective. Consider the injective entire function $$widetilde{f} colon z mapsto f(z) - f(0).$$ We see that $widetilde{f}(0) = 0$. Let $r > 0$, and consider the ball $B(0,r)$ of radius $r$ centered at the origin. By the Open Mapping Theorem, $widetilde{f}(B(0,r))$ is an open set containing the origin. By openness, there is some $epsilon > 0$ such that $B(0,epsilon) subset widetilde{f}(B(0,r))$. Since $widetilde{f}$ is injective, no point of $B(0,r)^c$ may be sent to any point of $B(0,epsilon)$. In other words, $$|z| geq r Rightarrow |widetilde{f}(z)| geq epsilon.$$
Now $widetilde{f}$ has a zero at the origin, so let's say this zero has order $k geq 1$. This means there is some entire function $g colon Bbb C rightarrow Bbb C$ such that $g(0) neq 0$ and $$widetilde{f}(z) = z^k g(z) forall z in Bbb C.$$
Since $widetilde{f}$ is injective, $widetilde{f}$ has no other zeroes apart from its zero at the origin. So $g$ has no zeroes. This means the function $F colon Bbb C rightarrow Bbb C$ defined by $$F(z) = frac{1}{g(z)} forall z in Bbb C$$ is entire and also has no zeroes. Write the Taylor expansion of $F$ at the origin as $$F(z) = sum_{n=0}^infty frac{F^{(n)}(0)}{n!}z^n.$$
Given any $R > r$, we can apply Cauchy's Estimates on the circle of radius $R$ to obtain bounds on the derivatives of $F$ at the origin. Indeed,
$$F^{(n)}(0) leq max_{|z| = R}|F(z)|frac{n! }{R^n}.$$ But we can apply the fact that $|widetilde{f}(z)| geq epsilon forall |z| geq r$ to notice that $$max_{|z| = R} |F(z)| = max_{|z| = R} frac{1}{|g(z)|} = max_{|z| = R} frac{|z|^k}{|widetilde{f}(z)|} leq frac{R^k}{epsilon}.$$ So if $n > k$, we have $$F^{(n)}(0) leq max_{|z| = R}|F(z)|frac{n!}{R^n} leq frac{R^k n!}{epsilon R^n} = frac{n!}{epsilon R^{n-k}} xrightarrow{R to infty} 0.$$
So we conclude from the Taylor expansion of $F$ that $$F(z) = sum_{n=0}^k frac{F^{(n)}(0)}{n!} z^n$$ is a polynomial of degree at most $k$. But as previously noted, $F$ has no zeroes. So $F equiv c$ for some $c inBbb C backslash{ 0}$! This means $g equiv c^{-1}$, and furthermore, $$widetilde{f}(z) = c^{-1}z^k forall z in Bbb C.$$
Supposing $k geq 2$, the polynomial $z^k - 1 in Bbb C[z]$ has at least two roots $xi_1, xi_2 in Bbb C$ which are certainly non-zero on account of the fact that $0^k neq 1$. The multiplicity of the root $xi_1$ is precisely $1$ because $$frac{d}{dz}Bigvert_{z=xi_1} z^k - 1 = k xi_1^{k-1} neq 0.$$ This allows us to conclude that $xi_1 neq xi_2$, and that $$widetilde{f}(xi_1) = c^{-1} xi_1^k = c^{-1} cdot 1 = c^{-1} xi_2^k = widetilde{f}(xi_2).$$ This result contradicts the fact that $widetilde{f}$ is injective. So $k=1$, and our final conclusion is that $$f(z) = widetilde{f}(z) + f(0) = c^{-1} z + f(0) forall z in Bbb C,$$ where for completeness's sake I recall that $c$ was necessarily non-zero.
answered Mar 3 '16 at 21:07
Open SeasonOpen Season
800518
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2
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Very nice! And you can even add that $c=1/(f(1)-f(0))$.
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– Unit
Mar 3 '16 at 21:15
add a comment |
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Here is a (longer) proof using very little of complex analysis. Assume that $f:mathbb{C}tomathbb{C}$ is holomorphic and injective. The function $f$ extends to a holomorphic map of the Riemann sphere to itself, $mathbb{CP}^1tomathbb{CP}^1$. Indeed, choose any $z_0$ such that $f'(z_0)neq 0$ (if it doesn't exist then $f$ is constant (of course from injectivity we know that actually $f'neq0$ everywhere)). Then a small neighbourhood $Uni z_0$ is mapped bijectively to a small neighbourhood $Vni f(z_0)$. The function $1/(f(z)-f(z_0))$ is therefore bounded in $mathbb{C}-U$, hence by Riemann removable singularity theorem it extends to a holomorphic function on $mathbb{CP}^1-U$, therefore $f$ extends to a holomorphic map $mathbb{CP}^1tomathbb{CP}^1$.
As an application of Liouville's theorem, any holomorphic map $F:mathbb{CP}^1tomathbb{CP}^1$ such $F(z)neqinfty$ for $zneqinfty$ is a polynomial. If we wish, we can also avoid Liuoville theorem and use some topology.
If the order of pole of $f$ at $infty$ is $>1$ then $f$ is not injective in the neighbourhood of $infty$. Hence there is $ainmathbb{C}$ such that $f-az$ is holomorphic $mathbb{CP}^1tomathbb{C}$, hence it's bounded (being a map from a compact space), hence it's a constant: if $f-az$ is not constant then it is a map $mathbb{CP}^1tomathbb{CP}^1$ which is of positive degree but which is not surjective (as it avoids $infty$).
answered Mar 29 '11 at 20:21
user8268user8268
17k12947
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add a comment |
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6 Answers
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6 Answers
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You can rule out polynomials of degree greater than $1$, because the derivative of such a polynomial will have a zero by the fundamental theorem of algebra, and a holomorphic function is $(n+1)$-to-$1$ near a zero of its derivative of order $n$.
To finish, you need to rule out entire functions that are not polynomials. If $f$ is such a function, then $f(1/z)$ has an essential singularity at $z=0$. To see that this implies that $f$ is not one-to-one, you could apply Picard's theorem as yoyo indicates. Or you could proceed as follows. By Casorati-Weierstrass, $f({z:|z|>n})$ is dense in $mathbb{C}$ for each positive integer $n$. By the open mapping theorem, the set is open. By Baire's theorem, $D=bigcap_n f({z:|z|>n})$ is dense in $mathbb{C}$. In particular, $D$ is not empty, and every element of $D$ has infinitely many preimage points under $f$.
I just realized that there is an easier way to apply Casorati-Weierstrass, with no need for Baire. If $f$ is entire and not a polynomial, then $f({z:|z|<1})$ is open, and $f({z:|z|>1})$ is dense. Therefore these sets have nonempty intersection. Every element of the intersection has at least $2$ preimage points.
edited May 27 '13 at 4:54
Michael Hardy
1
answered Mar 29 '11 at 17:18
Jonas MeyerJonas Meyer
41.1k6148258
$endgroup$
$begingroup$
Nice application of Baire. Thanks!
$endgroup$
– Digital Gal
Mar 29 '11 at 19:24
add a comment |
$begingroup$
You can rule out polynomials of degree greater than $1$, because the derivative of such a polynomial will have a zero by the fundamental theorem of algebra, and a holomorphic function is $(n+1)$-to-$1$ near a zero of its derivative of order $n$.
To finish, you need to rule out entire functions that are not polynomials. If $f$ is such a function, then $f(1/z)$ has an essential singularity at $z=0$. To see that this implies that $f$ is not one-to-one, you could apply Picard's theorem as yoyo indicates. Or you could proceed as follows. By Casorati-Weierstrass, $f({z:|z|>n})$ is dense in $mathbb{C}$ for each positive integer $n$. By the open mapping theorem, the set is open. By Baire's theorem, $D=bigcap_n f({z:|z|>n})$ is dense in $mathbb{C}$. In particular, $D$ is not empty, and every element of $D$ has infinitely many preimage points under $f$.
I just realized that there is an easier way to apply Casorati-Weierstrass, with no need for Baire. If $f$ is entire and not a polynomial, then $f({z:|z|<1})$ is open, and $f({z:|z|>1})$ is dense. Therefore these sets have nonempty intersection. Every element of the intersection has at least $2$ preimage points.
edited May 27 '13 at 4:54
Michael Hardy
1
answered Mar 29 '11 at 17:18
Jonas MeyerJonas Meyer
41.1k6148258
$endgroup$
$begingroup$
Nice application of Baire. Thanks!
$endgroup$
– Digital Gal
Mar 29 '11 at 19:24
add a comment |
$begingroup$
You can rule out polynomials of degree greater than $1$, because the derivative of such a polynomial will have a zero by the fundamental theorem of algebra, and a holomorphic function is $(n+1)$-to-$1$ near a zero of its derivative of order $n$.
To finish, you need to rule out entire functions that are not polynomials. If $f$ is such a function, then $f(1/z)$ has an essential singularity at $z=0$. To see that this implies that $f$ is not one-to-one, you could apply Picard's theorem as yoyo indicates. Or you could proceed as follows. By Casorati-Weierstrass, $f({z:|z|>n})$ is dense in $mathbb{C}$ for each positive integer $n$. By the open mapping theorem, the set is open. By Baire's theorem, $D=bigcap_n f({z:|z|>n})$ is dense in $mathbb{C}$. In particular, $D$ is not empty, and every element of $D$ has infinitely many preimage points under $f$.
I just realized that there is an easier way to apply Casorati-Weierstrass, with no need for Baire. If $f$ is entire and not a polynomial, then $f({z:|z|<1})$ is open, and $f({z:|z|>1})$ is dense. Therefore these sets have nonempty intersection. Every element of the intersection has at least $2$ preimage points.
edited May 27 '13 at 4:54
Michael Hardy
1
answered Mar 29 '11 at 17:18
Jonas MeyerJonas Meyer
41.1k6148258
$endgroup$
You can rule out polynomials of degree greater than $1$, because the derivative of such a polynomial will have a zero by the fundamental theorem of algebra, and a holomorphic function is $(n+1)$-to-$1$ near a zero of its derivative of order $n$.
To finish, you need to rule out entire functions that are not polynomials. If $f$ is such a function, then $f(1/z)$ has an essential singularity at $z=0$. To see that this implies that $f$ is not one-to-one, you could apply Picard's theorem as yoyo indicates. Or you could proceed as follows. By Casorati-Weierstrass, $f({z:|z|>n})$ is dense in $mathbb{C}$ for each positive integer $n$. By the open mapping theorem, the set is open. By Baire's theorem, $D=bigcap_n f({z:|z|>n})$ is dense in $mathbb{C}$. In particular, $D$ is not empty, and every element of $D$ has infinitely many preimage points under $f$.
I just realized that there is an easier way to apply Casorati-Weierstrass, with no need for Baire. If $f$ is entire and not a polynomial, then $f({z:|z|<1})$ is open, and $f({z:|z|>1})$ is dense. Therefore these sets have nonempty intersection. Every element of the intersection has at least $2$ preimage points.
edited May 27 '13 at 4:54
Michael Hardy
1
answered Mar 29 '11 at 17:18
Jonas MeyerJonas Meyer
41.1k6148258
edited May 27 '13 at 4:54
Michael Hardy
1
edited May 27 '13 at 4:54
Michael Hardy
1
edited May 27 '13 at 4:54
Michael Hardy
1
1
answered Mar 29 '11 at 17:18
Jonas MeyerJonas Meyer
41.1k6148258
answered Mar 29 '11 at 17:18
Jonas MeyerJonas Meyer
41.1k6148258
answered Mar 29 '11 at 17:18
Jonas MeyerJonas Meyer
41.1k6148258
41.1k6148258
$begingroup$
Nice application of Baire. Thanks!
$endgroup$
– Digital Gal
Mar 29 '11 at 19:24
add a comment |
$begingroup$
Nice application of Baire. Thanks!
$endgroup$
– Digital Gal
Mar 29 '11 at 19:24
$begingroup$
Nice application of Baire. Thanks!
$endgroup$
– Digital Gal
Mar 29 '11 at 19:24
$begingroup$
Nice application of Baire. Thanks!
$endgroup$
– Digital Gal
Mar 29 '11 at 19:24
add a comment |
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By shifting $z$, without loss of generality you can assume $f(0) = 0$. By the open mapping theorem, $f(z)$ maps some open set $U$ containing $0$ to another one, call it $V$. Since $f(z)$ is to be one-to-one, $f(z)$ can't map any $z$ outside of $U$ to $V$. Thus ${1 over f(z)}$ is bounded outside of $U$. Therefore ${z over f(z)}$ is an entire function that grows
no faster than linearly: $|{z over f(z)}| < A|z| + B$ for some $A$ and $B$.
It's easy from here to show that $g(z) = {z over f(z)}$ is linear; for any $z_0$
${g(z) - g(z_0) over z - z_0}$ must be bounded and therefore is a constant by Liouville's theorem. So ${z over f(z)} = c_1z + c_2$ for some $c_1$ and $c_2$. Hence $f(z) = {zover c_1z + c_2}$. Since $f(z)$ has no poles and is nonconstant, $c_1$ must be zero and $c_2$ nonzero. We conclude that $f(z) = {1 over c_2} z$.
answered Mar 29 '11 at 22:16
ZarraxZarrax
35.7k250104
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Why f(z) has simple zero at z=0?
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– nicksohn
Dec 30 '15 at 10:33
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Nice proof. In the beginning, you say wlog wma $f(z)=0$. Maybe you mean $f(0)=0$?
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– perlman
Oct 12 '17 at 5:55
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@nicksohn That follows from $f(z)$ being one to one
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– Zarrax
Oct 12 '17 at 17:40
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@MathewJames yes, thanks, corrected it.
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– Zarrax
Oct 12 '17 at 17:40
add a comment |
$begingroup$
By shifting $z$, without loss of generality you can assume $f(0) = 0$. By the open mapping theorem, $f(z)$ maps some open set $U$ containing $0$ to another one, call it $V$. Since $f(z)$ is to be one-to-one, $f(z)$ can't map any $z$ outside of $U$ to $V$. Thus ${1 over f(z)}$ is bounded outside of $U$. Therefore ${z over f(z)}$ is an entire function that grows
no faster than linearly: $|{z over f(z)}| < A|z| + B$ for some $A$ and $B$.
It's easy from here to show that $g(z) = {z over f(z)}$ is linear; for any $z_0$
${g(z) - g(z_0) over z - z_0}$ must be bounded and therefore is a constant by Liouville's theorem. So ${z over f(z)} = c_1z + c_2$ for some $c_1$ and $c_2$. Hence $f(z) = {zover c_1z + c_2}$. Since $f(z)$ has no poles and is nonconstant, $c_1$ must be zero and $c_2$ nonzero. We conclude that $f(z) = {1 over c_2} z$.
answered Mar 29 '11 at 22:16
ZarraxZarrax
35.7k250104
$endgroup$
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Why f(z) has simple zero at z=0?
$endgroup$
– nicksohn
Dec 30 '15 at 10:33
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Nice proof. In the beginning, you say wlog wma $f(z)=0$. Maybe you mean $f(0)=0$?
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– perlman
Oct 12 '17 at 5:55
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@nicksohn That follows from $f(z)$ being one to one
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– Zarrax
Oct 12 '17 at 17:40
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@MathewJames yes, thanks, corrected it.
$endgroup$
– Zarrax
Oct 12 '17 at 17:40
add a comment |
$begingroup$
By shifting $z$, without loss of generality you can assume $f(0) = 0$. By the open mapping theorem, $f(z)$ maps some open set $U$ containing $0$ to another one, call it $V$. Since $f(z)$ is to be one-to-one, $f(z)$ can't map any $z$ outside of $U$ to $V$. Thus ${1 over f(z)}$ is bounded outside of $U$. Therefore ${z over f(z)}$ is an entire function that grows
no faster than linearly: $|{z over f(z)}| < A|z| + B$ for some $A$ and $B$.
It's easy from here to show that $g(z) = {z over f(z)}$ is linear; for any $z_0$
${g(z) - g(z_0) over z - z_0}$ must be bounded and therefore is a constant by Liouville's theorem. So ${z over f(z)} = c_1z + c_2$ for some $c_1$ and $c_2$. Hence $f(z) = {zover c_1z + c_2}$. Since $f(z)$ has no poles and is nonconstant, $c_1$ must be zero and $c_2$ nonzero. We conclude that $f(z) = {1 over c_2} z$.
answered Mar 29 '11 at 22:16
ZarraxZarrax
35.7k250104
$endgroup$
By shifting $z$, without loss of generality you can assume $f(0) = 0$. By the open mapping theorem, $f(z)$ maps some open set $U$ containing $0$ to another one, call it $V$. Since $f(z)$ is to be one-to-one, $f(z)$ can't map any $z$ outside of $U$ to $V$. Thus ${1 over f(z)}$ is bounded outside of $U$. Therefore ${z over f(z)}$ is an entire function that grows
no faster than linearly: $|{z over f(z)}| < A|z| + B$ for some $A$ and $B$.
It's easy from here to show that $g(z) = {z over f(z)}$ is linear; for any $z_0$
${g(z) - g(z_0) over z - z_0}$ must be bounded and therefore is a constant by Liouville's theorem. So ${z over f(z)} = c_1z + c_2$ for some $c_1$ and $c_2$. Hence $f(z) = {zover c_1z + c_2}$. Since $f(z)$ has no poles and is nonconstant, $c_1$ must be zero and $c_2$ nonzero. We conclude that $f(z) = {1 over c_2} z$.
answered Mar 29 '11 at 22:16
ZarraxZarrax
35.7k250104
edited Oct 12 '17 at 17:40
answered Mar 29 '11 at 22:16
ZarraxZarrax
35.7k250104
answered Mar 29 '11 at 22:16
ZarraxZarrax
35.7k250104
answered Mar 29 '11 at 22:16
ZarraxZarrax
35.7k250104
35.7k250104
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Why f(z) has simple zero at z=0?
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– nicksohn
Dec 30 '15 at 10:33
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Nice proof. In the beginning, you say wlog wma $f(z)=0$. Maybe you mean $f(0)=0$?
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– perlman
Oct 12 '17 at 5:55
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@nicksohn That follows from $f(z)$ being one to one
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– Zarrax
Oct 12 '17 at 17:40
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@MathewJames yes, thanks, corrected it.
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– Zarrax
Oct 12 '17 at 17:40
add a comment |
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Why f(z) has simple zero at z=0?
$endgroup$
– nicksohn
Dec 30 '15 at 10:33
$begingroup$
Nice proof. In the beginning, you say wlog wma $f(z)=0$. Maybe you mean $f(0)=0$?
$endgroup$
– perlman
Oct 12 '17 at 5:55
$begingroup$
@nicksohn That follows from $f(z)$ being one to one
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– Zarrax
Oct 12 '17 at 17:40
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@MathewJames yes, thanks, corrected it.
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– Zarrax
Oct 12 '17 at 17:40
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Why f(z) has simple zero at z=0?
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– nicksohn
Dec 30 '15 at 10:33
$begingroup$
Why f(z) has simple zero at z=0?
$endgroup$
– nicksohn
Dec 30 '15 at 10:33
$begingroup$
Nice proof. In the beginning, you say wlog wma $f(z)=0$. Maybe you mean $f(0)=0$?
$endgroup$
– perlman
Oct 12 '17 at 5:55
$begingroup$
Nice proof. In the beginning, you say wlog wma $f(z)=0$. Maybe you mean $f(0)=0$?
$endgroup$
– perlman
Oct 12 '17 at 5:55
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@nicksohn That follows from $f(z)$ being one to one
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– Zarrax
Oct 12 '17 at 17:40
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@nicksohn That follows from $f(z)$ being one to one
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– Zarrax
Oct 12 '17 at 17:40
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@MathewJames yes, thanks, corrected it.
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– Zarrax
Oct 12 '17 at 17:40
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@MathewJames yes, thanks, corrected it.
$endgroup$
– Zarrax
Oct 12 '17 at 17:40
add a comment |
$begingroup$
Let $f:mathbb Ctomathbb C$ entire and injective. Let $U=f(mathbb C)$. $U$ is an open subset of the plane.
$U$ is simply connected: indeed, to check this it is enough to show that the integral of every analytic function on $U$ along every closed curved in $U$ is zero, and you can do this by "changing variables using $f$".
Next, if $Usubsetneqmathbb C$, from Riemann's theorem we know that there is an biholomorphic map $Uto D$, with $D$ the unit disc. Composing with $f$, we get a biholomorphic map $mathbb Cto D$, and this is impossible. We see then that $f$ is in fact bijective and, in fact, an homeomorphism. Composing with a translation, we can assume that $f(0)=0$.
Using this, one can see that the function $1/f(z)$ is bounded at $infty$ and has a simple pole at $0$, so $g(z)=z/f(z)$ is entire and bounded by a function of the form $cz$ for some constant $c$. Using Cauchy's estimates for the Taylor coefficients of $g$, we see that $g$ is a polynomial of very low degree. Translating this to information about $f$, we can conclude what we want.
(This avoids Picard but uses Riemann... :( )
answered Mar 29 '11 at 17:28
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
112k7159291
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How do we know that the pole at 0 is of order 1, from the fact that f(0)=0? What if 0 was zero of order n>1 for f(z)...thanks, @Mariano Suárez-Alvarez.
$endgroup$
– User001
Dec 18 '14 at 6:18
add a comment |
$begingroup$
Let $f:mathbb Ctomathbb C$ entire and injective. Let $U=f(mathbb C)$. $U$ is an open subset of the plane.
$U$ is simply connected: indeed, to check this it is enough to show that the integral of every analytic function on $U$ along every closed curved in $U$ is zero, and you can do this by "changing variables using $f$".
Next, if $Usubsetneqmathbb C$, from Riemann's theorem we know that there is an biholomorphic map $Uto D$, with $D$ the unit disc. Composing with $f$, we get a biholomorphic map $mathbb Cto D$, and this is impossible. We see then that $f$ is in fact bijective and, in fact, an homeomorphism. Composing with a translation, we can assume that $f(0)=0$.
Using this, one can see that the function $1/f(z)$ is bounded at $infty$ and has a simple pole at $0$, so $g(z)=z/f(z)$ is entire and bounded by a function of the form $cz$ for some constant $c$. Using Cauchy's estimates for the Taylor coefficients of $g$, we see that $g$ is a polynomial of very low degree. Translating this to information about $f$, we can conclude what we want.
(This avoids Picard but uses Riemann... :( )
answered Mar 29 '11 at 17:28
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
112k7159291
$endgroup$
$begingroup$
How do we know that the pole at 0 is of order 1, from the fact that f(0)=0? What if 0 was zero of order n>1 for f(z)...thanks, @Mariano Suárez-Alvarez.
$endgroup$
– User001
Dec 18 '14 at 6:18
add a comment |
$begingroup$
Let $f:mathbb Ctomathbb C$ entire and injective. Let $U=f(mathbb C)$. $U$ is an open subset of the plane.
$U$ is simply connected: indeed, to check this it is enough to show that the integral of every analytic function on $U$ along every closed curved in $U$ is zero, and you can do this by "changing variables using $f$".
Next, if $Usubsetneqmathbb C$, from Riemann's theorem we know that there is an biholomorphic map $Uto D$, with $D$ the unit disc. Composing with $f$, we get a biholomorphic map $mathbb Cto D$, and this is impossible. We see then that $f$ is in fact bijective and, in fact, an homeomorphism. Composing with a translation, we can assume that $f(0)=0$.
Using this, one can see that the function $1/f(z)$ is bounded at $infty$ and has a simple pole at $0$, so $g(z)=z/f(z)$ is entire and bounded by a function of the form $cz$ for some constant $c$. Using Cauchy's estimates for the Taylor coefficients of $g$, we see that $g$ is a polynomial of very low degree. Translating this to information about $f$, we can conclude what we want.
(This avoids Picard but uses Riemann... :( )
answered Mar 29 '11 at 17:28
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
112k7159291
$endgroup$
Let $f:mathbb Ctomathbb C$ entire and injective. Let $U=f(mathbb C)$. $U$ is an open subset of the plane.
$U$ is simply connected: indeed, to check this it is enough to show that the integral of every analytic function on $U$ along every closed curved in $U$ is zero, and you can do this by "changing variables using $f$".
Next, if $Usubsetneqmathbb C$, from Riemann's theorem we know that there is an biholomorphic map $Uto D$, with $D$ the unit disc. Composing with $f$, we get a biholomorphic map $mathbb Cto D$, and this is impossible. We see then that $f$ is in fact bijective and, in fact, an homeomorphism. Composing with a translation, we can assume that $f(0)=0$.
Using this, one can see that the function $1/f(z)$ is bounded at $infty$ and has a simple pole at $0$, so $g(z)=z/f(z)$ is entire and bounded by a function of the form $cz$ for some constant $c$. Using Cauchy's estimates for the Taylor coefficients of $g$, we see that $g$ is a polynomial of very low degree. Translating this to information about $f$, we can conclude what we want.
(This avoids Picard but uses Riemann... :( )
answered Mar 29 '11 at 17:28
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
112k7159291
edited Mar 29 '11 at 17:36
answered Mar 29 '11 at 17:28
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
112k7159291
answered Mar 29 '11 at 17:28
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
112k7159291
answered Mar 29 '11 at 17:28
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
112k7159291
112k7159291
$begingroup$
How do we know that the pole at 0 is of order 1, from the fact that f(0)=0? What if 0 was zero of order n>1 for f(z)...thanks, @Mariano Suárez-Alvarez.
$endgroup$
– User001
Dec 18 '14 at 6:18
add a comment |
$begingroup$
How do we know that the pole at 0 is of order 1, from the fact that f(0)=0? What if 0 was zero of order n>1 for f(z)...thanks, @Mariano Suárez-Alvarez.
$endgroup$
– User001
Dec 18 '14 at 6:18
$begingroup$
How do we know that the pole at 0 is of order 1, from the fact that f(0)=0? What if 0 was zero of order n>1 for f(z)...thanks, @Mariano Suárez-Alvarez.
$endgroup$
– User001
Dec 18 '14 at 6:18
$begingroup$
How do we know that the pole at 0 is of order 1, from the fact that f(0)=0? What if 0 was zero of order n>1 for f(z)...thanks, @Mariano Suárez-Alvarez.
$endgroup$
– User001
Dec 18 '14 at 6:18
add a comment |
$begingroup$
I'll give the "usual" proof.
Note that by Little Picard, $f$ misses at most one point; but it is a homeomorphism onto its image, and the plane minus a point is not simply connected. Thus $f$ is onto $mathbb{C}$, and hence bijective. Then $f$ has a holomorphic inverse, which is enough to imply $f$ is proper, that is, the pre-image of a compact set is compact. This in turn implies
$$ lim_{zrightarrowinfty} f(z)=infty,$$
and thus if we define $f(infty)=infty$, $f$ becomes a Möbius transformation of the Riemann sphere. So $f$ has the form
$f(z) = frac{az+b}{cz+d},$
and it is easy to see that if $f$ is entire on $mathbb{C}$, then $c=0$.
edited Sep 29 '13 at 9:47
azimut
16.6k1052101
$endgroup$
$begingroup$
I should also add that I believe there is a proof using just Little Picard and the Cauchy-Riemann equations. I just forget all the details :-). Basically you show $f$ is a Euclidean similarity on $mathbb{R}^2$, so it has the form $kAz+b$, where $A$ is $2times2$ orthogonal. Since $f$ is holomorphic, $Ain SO(2)$. Using the well known form of such matrices, you then show $A$ is rotation by some angle $theta$, so $f(z)=ke^{itheta}z+b$.
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– user641
Mar 29 '11 at 20:36
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Why is it that $c=0$ if $f$ is entire?
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– user225477
Feb 16 '17 at 18:33
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@Zermelo's_Choice, I think otherwise $f$ would have a pole at the point $z$ such that $cz+d=0$.
$endgroup$
– user135520
Apr 27 '18 at 16:43
add a comment |
$begingroup$
I'll give the "usual" proof.
Note that by Little Picard, $f$ misses at most one point; but it is a homeomorphism onto its image, and the plane minus a point is not simply connected. Thus $f$ is onto $mathbb{C}$, and hence bijective. Then $f$ has a holomorphic inverse, which is enough to imply $f$ is proper, that is, the pre-image of a compact set is compact. This in turn implies
$$ lim_{zrightarrowinfty} f(z)=infty,$$
and thus if we define $f(infty)=infty$, $f$ becomes a Möbius transformation of the Riemann sphere. So $f$ has the form
$f(z) = frac{az+b}{cz+d},$
and it is easy to see that if $f$ is entire on $mathbb{C}$, then $c=0$.
edited Sep 29 '13 at 9:47
azimut
16.6k1052101
$endgroup$
$begingroup$
I should also add that I believe there is a proof using just Little Picard and the Cauchy-Riemann equations. I just forget all the details :-). Basically you show $f$ is a Euclidean similarity on $mathbb{R}^2$, so it has the form $kAz+b$, where $A$ is $2times2$ orthogonal. Since $f$ is holomorphic, $Ain SO(2)$. Using the well known form of such matrices, you then show $A$ is rotation by some angle $theta$, so $f(z)=ke^{itheta}z+b$.
$endgroup$
– user641
Mar 29 '11 at 20:36
$begingroup$
Why is it that $c=0$ if $f$ is entire?
$endgroup$
– user225477
Feb 16 '17 at 18:33
$begingroup$
@Zermelo's_Choice, I think otherwise $f$ would have a pole at the point $z$ such that $cz+d=0$.
$endgroup$
– user135520
Apr 27 '18 at 16:43
add a comment |
$begingroup$
I'll give the "usual" proof.
Note that by Little Picard, $f$ misses at most one point; but it is a homeomorphism onto its image, and the plane minus a point is not simply connected. Thus $f$ is onto $mathbb{C}$, and hence bijective. Then $f$ has a holomorphic inverse, which is enough to imply $f$ is proper, that is, the pre-image of a compact set is compact. This in turn implies
$$ lim_{zrightarrowinfty} f(z)=infty,$$
and thus if we define $f(infty)=infty$, $f$ becomes a Möbius transformation of the Riemann sphere. So $f$ has the form
$f(z) = frac{az+b}{cz+d},$
and it is easy to see that if $f$ is entire on $mathbb{C}$, then $c=0$.
edited Sep 29 '13 at 9:47
azimut
16.6k1052101
$endgroup$
I'll give the "usual" proof.
Note that by Little Picard, $f$ misses at most one point; but it is a homeomorphism onto its image, and the plane minus a point is not simply connected. Thus $f$ is onto $mathbb{C}$, and hence bijective. Then $f$ has a holomorphic inverse, which is enough to imply $f$ is proper, that is, the pre-image of a compact set is compact. This in turn implies
$$ lim_{zrightarrowinfty} f(z)=infty,$$
and thus if we define $f(infty)=infty$, $f$ becomes a Möbius transformation of the Riemann sphere. So $f$ has the form
$f(z) = frac{az+b}{cz+d},$
and it is easy to see that if $f$ is entire on $mathbb{C}$, then $c=0$.
edited Sep 29 '13 at 9:47
azimut
16.6k1052101
edited Sep 29 '13 at 9:47
azimut
16.6k1052101
edited Sep 29 '13 at 9:47
azimut
16.6k1052101
edited Sep 29 '13 at 9:47
azimut
16.6k1052101
16.6k1052101
answered Mar 29 '11 at 20:19
user641
$begingroup$
I should also add that I believe there is a proof using just Little Picard and the Cauchy-Riemann equations. I just forget all the details :-). Basically you show $f$ is a Euclidean similarity on $mathbb{R}^2$, so it has the form $kAz+b$, where $A$ is $2times2$ orthogonal. Since $f$ is holomorphic, $Ain SO(2)$. Using the well known form of such matrices, you then show $A$ is rotation by some angle $theta$, so $f(z)=ke^{itheta}z+b$.
$endgroup$
– user641
Mar 29 '11 at 20:36
$begingroup$
Why is it that $c=0$ if $f$ is entire?
$endgroup$
– user225477
Feb 16 '17 at 18:33
$begingroup$
@Zermelo's_Choice, I think otherwise $f$ would have a pole at the point $z$ such that $cz+d=0$.
$endgroup$
– user135520
Apr 27 '18 at 16:43
add a comment |
$begingroup$
I should also add that I believe there is a proof using just Little Picard and the Cauchy-Riemann equations. I just forget all the details :-). Basically you show $f$ is a Euclidean similarity on $mathbb{R}^2$, so it has the form $kAz+b$, where $A$ is $2times2$ orthogonal. Since $f$ is holomorphic, $Ain SO(2)$. Using the well known form of such matrices, you then show $A$ is rotation by some angle $theta$, so $f(z)=ke^{itheta}z+b$.
$endgroup$
– user641
Mar 29 '11 at 20:36
$begingroup$
Why is it that $c=0$ if $f$ is entire?
$endgroup$
– user225477
Feb 16 '17 at 18:33
$begingroup$
@Zermelo's_Choice, I think otherwise $f$ would have a pole at the point $z$ such that $cz+d=0$.
$endgroup$
– user135520
Apr 27 '18 at 16:43
$begingroup$
I should also add that I believe there is a proof using just Little Picard and the Cauchy-Riemann equations. I just forget all the details :-). Basically you show $f$ is a Euclidean similarity on $mathbb{R}^2$, so it has the form $kAz+b$, where $A$ is $2times2$ orthogonal. Since $f$ is holomorphic, $Ain SO(2)$. Using the well known form of such matrices, you then show $A$ is rotation by some angle $theta$, so $f(z)=ke^{itheta}z+b$.
$endgroup$
– user641
Mar 29 '11 at 20:36
$begingroup$
I should also add that I believe there is a proof using just Little Picard and the Cauchy-Riemann equations. I just forget all the details :-). Basically you show $f$ is a Euclidean similarity on $mathbb{R}^2$, so it has the form $kAz+b$, where $A$ is $2times2$ orthogonal. Since $f$ is holomorphic, $Ain SO(2)$. Using the well known form of such matrices, you then show $A$ is rotation by some angle $theta$, so $f(z)=ke^{itheta}z+b$.
$endgroup$
– user641
Mar 29 '11 at 20:36
$begingroup$
Why is it that $c=0$ if $f$ is entire?
$endgroup$
– user225477
Feb 16 '17 at 18:33
$begingroup$
Why is it that $c=0$ if $f$ is entire?
$endgroup$
– user225477
Feb 16 '17 at 18:33
$begingroup$
@Zermelo's_Choice, I think otherwise $f$ would have a pole at the point $z$ such that $cz+d=0$.
$endgroup$
– user135520
Apr 27 '18 at 16:43
$begingroup$
@Zermelo's_Choice, I think otherwise $f$ would have a pole at the point $z$ such that $cz+d=0$.
$endgroup$
– user135520
Apr 27 '18 at 16:43
add a comment |
$begingroup$
This proof uses only the Open Mapping Theorem and Cauchy's Estimates.
Suppose $f$ is entire and injective. Consider the injective entire function $$widetilde{f} colon z mapsto f(z) - f(0).$$ We see that $widetilde{f}(0) = 0$. Let $r > 0$, and consider the ball $B(0,r)$ of radius $r$ centered at the origin. By the Open Mapping Theorem, $widetilde{f}(B(0,r))$ is an open set containing the origin. By openness, there is some $epsilon > 0$ such that $B(0,epsilon) subset widetilde{f}(B(0,r))$. Since $widetilde{f}$ is injective, no point of $B(0,r)^c$ may be sent to any point of $B(0,epsilon)$. In other words, $$|z| geq r Rightarrow |widetilde{f}(z)| geq epsilon.$$
Now $widetilde{f}$ has a zero at the origin, so let's say this zero has order $k geq 1$. This means there is some entire function $g colon Bbb C rightarrow Bbb C$ such that $g(0) neq 0$ and $$widetilde{f}(z) = z^k g(z) forall z in Bbb C.$$
Since $widetilde{f}$ is injective, $widetilde{f}$ has no other zeroes apart from its zero at the origin. So $g$ has no zeroes. This means the function $F colon Bbb C rightarrow Bbb C$ defined by $$F(z) = frac{1}{g(z)} forall z in Bbb C$$ is entire and also has no zeroes. Write the Taylor expansion of $F$ at the origin as $$F(z) = sum_{n=0}^infty frac{F^{(n)}(0)}{n!}z^n.$$
Given any $R > r$, we can apply Cauchy's Estimates on the circle of radius $R$ to obtain bounds on the derivatives of $F$ at the origin. Indeed,
$$F^{(n)}(0) leq max_{|z| = R}|F(z)|frac{n! }{R^n}.$$ But we can apply the fact that $|widetilde{f}(z)| geq epsilon forall |z| geq r$ to notice that $$max_{|z| = R} |F(z)| = max_{|z| = R} frac{1}{|g(z)|} = max_{|z| = R} frac{|z|^k}{|widetilde{f}(z)|} leq frac{R^k}{epsilon}.$$ So if $n > k$, we have $$F^{(n)}(0) leq max_{|z| = R}|F(z)|frac{n!}{R^n} leq frac{R^k n!}{epsilon R^n} = frac{n!}{epsilon R^{n-k}} xrightarrow{R to infty} 0.$$
So we conclude from the Taylor expansion of $F$ that $$F(z) = sum_{n=0}^k frac{F^{(n)}(0)}{n!} z^n$$ is a polynomial of degree at most $k$. But as previously noted, $F$ has no zeroes. So $F equiv c$ for some $c inBbb C backslash{ 0}$! This means $g equiv c^{-1}$, and furthermore, $$widetilde{f}(z) = c^{-1}z^k forall z in Bbb C.$$
Supposing $k geq 2$, the polynomial $z^k - 1 in Bbb C[z]$ has at least two roots $xi_1, xi_2 in Bbb C$ which are certainly non-zero on account of the fact that $0^k neq 1$. The multiplicity of the root $xi_1$ is precisely $1$ because $$frac{d}{dz}Bigvert_{z=xi_1} z^k - 1 = k xi_1^{k-1} neq 0.$$ This allows us to conclude that $xi_1 neq xi_2$, and that $$widetilde{f}(xi_1) = c^{-1} xi_1^k = c^{-1} cdot 1 = c^{-1} xi_2^k = widetilde{f}(xi_2).$$ This result contradicts the fact that $widetilde{f}$ is injective. So $k=1$, and our final conclusion is that $$f(z) = widetilde{f}(z) + f(0) = c^{-1} z + f(0) forall z in Bbb C,$$ where for completeness's sake I recall that $c$ was necessarily non-zero.
answered Mar 3 '16 at 21:07
Open SeasonOpen Season
800518
$endgroup$
2
$begingroup$
Very nice! And you can even add that $c=1/(f(1)-f(0))$.
$endgroup$
– Unit
Mar 3 '16 at 21:15
add a comment |
$begingroup$
This proof uses only the Open Mapping Theorem and Cauchy's Estimates.
Suppose $f$ is entire and injective. Consider the injective entire function $$widetilde{f} colon z mapsto f(z) - f(0).$$ We see that $widetilde{f}(0) = 0$. Let $r > 0$, and consider the ball $B(0,r)$ of radius $r$ centered at the origin. By the Open Mapping Theorem, $widetilde{f}(B(0,r))$ is an open set containing the origin. By openness, there is some $epsilon > 0$ such that $B(0,epsilon) subset widetilde{f}(B(0,r))$. Since $widetilde{f}$ is injective, no point of $B(0,r)^c$ may be sent to any point of $B(0,epsilon)$. In other words, $$|z| geq r Rightarrow |widetilde{f}(z)| geq epsilon.$$
Now $widetilde{f}$ has a zero at the origin, so let's say this zero has order $k geq 1$. This means there is some entire function $g colon Bbb C rightarrow Bbb C$ such that $g(0) neq 0$ and $$widetilde{f}(z) = z^k g(z) forall z in Bbb C.$$
Since $widetilde{f}$ is injective, $widetilde{f}$ has no other zeroes apart from its zero at the origin. So $g$ has no zeroes. This means the function $F colon Bbb C rightarrow Bbb C$ defined by $$F(z) = frac{1}{g(z)} forall z in Bbb C$$ is entire and also has no zeroes. Write the Taylor expansion of $F$ at the origin as $$F(z) = sum_{n=0}^infty frac{F^{(n)}(0)}{n!}z^n.$$
Given any $R > r$, we can apply Cauchy's Estimates on the circle of radius $R$ to obtain bounds on the derivatives of $F$ at the origin. Indeed,
$$F^{(n)}(0) leq max_{|z| = R}|F(z)|frac{n! }{R^n}.$$ But we can apply the fact that $|widetilde{f}(z)| geq epsilon forall |z| geq r$ to notice that $$max_{|z| = R} |F(z)| = max_{|z| = R} frac{1}{|g(z)|} = max_{|z| = R} frac{|z|^k}{|widetilde{f}(z)|} leq frac{R^k}{epsilon}.$$ So if $n > k$, we have $$F^{(n)}(0) leq max_{|z| = R}|F(z)|frac{n!}{R^n} leq frac{R^k n!}{epsilon R^n} = frac{n!}{epsilon R^{n-k}} xrightarrow{R to infty} 0.$$
So we conclude from the Taylor expansion of $F$ that $$F(z) = sum_{n=0}^k frac{F^{(n)}(0)}{n!} z^n$$ is a polynomial of degree at most $k$. But as previously noted, $F$ has no zeroes. So $F equiv c$ for some $c inBbb C backslash{ 0}$! This means $g equiv c^{-1}$, and furthermore, $$widetilde{f}(z) = c^{-1}z^k forall z in Bbb C.$$
Supposing $k geq 2$, the polynomial $z^k - 1 in Bbb C[z]$ has at least two roots $xi_1, xi_2 in Bbb C$ which are certainly non-zero on account of the fact that $0^k neq 1$. The multiplicity of the root $xi_1$ is precisely $1$ because $$frac{d}{dz}Bigvert_{z=xi_1} z^k - 1 = k xi_1^{k-1} neq 0.$$ This allows us to conclude that $xi_1 neq xi_2$, and that $$widetilde{f}(xi_1) = c^{-1} xi_1^k = c^{-1} cdot 1 = c^{-1} xi_2^k = widetilde{f}(xi_2).$$ This result contradicts the fact that $widetilde{f}$ is injective. So $k=1$, and our final conclusion is that $$f(z) = widetilde{f}(z) + f(0) = c^{-1} z + f(0) forall z in Bbb C,$$ where for completeness's sake I recall that $c$ was necessarily non-zero.
answered Mar 3 '16 at 21:07
Open SeasonOpen Season
800518
$endgroup$
2
$begingroup$
Very nice! And you can even add that $c=1/(f(1)-f(0))$.
$endgroup$
– Unit
Mar 3 '16 at 21:15
add a comment |
$begingroup$
This proof uses only the Open Mapping Theorem and Cauchy's Estimates.
Suppose $f$ is entire and injective. Consider the injective entire function $$widetilde{f} colon z mapsto f(z) - f(0).$$ We see that $widetilde{f}(0) = 0$. Let $r > 0$, and consider the ball $B(0,r)$ of radius $r$ centered at the origin. By the Open Mapping Theorem, $widetilde{f}(B(0,r))$ is an open set containing the origin. By openness, there is some $epsilon > 0$ such that $B(0,epsilon) subset widetilde{f}(B(0,r))$. Since $widetilde{f}$ is injective, no point of $B(0,r)^c$ may be sent to any point of $B(0,epsilon)$. In other words, $$|z| geq r Rightarrow |widetilde{f}(z)| geq epsilon.$$
Now $widetilde{f}$ has a zero at the origin, so let's say this zero has order $k geq 1$. This means there is some entire function $g colon Bbb C rightarrow Bbb C$ such that $g(0) neq 0$ and $$widetilde{f}(z) = z^k g(z) forall z in Bbb C.$$
Since $widetilde{f}$ is injective, $widetilde{f}$ has no other zeroes apart from its zero at the origin. So $g$ has no zeroes. This means the function $F colon Bbb C rightarrow Bbb C$ defined by $$F(z) = frac{1}{g(z)} forall z in Bbb C$$ is entire and also has no zeroes. Write the Taylor expansion of $F$ at the origin as $$F(z) = sum_{n=0}^infty frac{F^{(n)}(0)}{n!}z^n.$$
Given any $R > r$, we can apply Cauchy's Estimates on the circle of radius $R$ to obtain bounds on the derivatives of $F$ at the origin. Indeed,
$$F^{(n)}(0) leq max_{|z| = R}|F(z)|frac{n! }{R^n}.$$ But we can apply the fact that $|widetilde{f}(z)| geq epsilon forall |z| geq r$ to notice that $$max_{|z| = R} |F(z)| = max_{|z| = R} frac{1}{|g(z)|} = max_{|z| = R} frac{|z|^k}{|widetilde{f}(z)|} leq frac{R^k}{epsilon}.$$ So if $n > k$, we have $$F^{(n)}(0) leq max_{|z| = R}|F(z)|frac{n!}{R^n} leq frac{R^k n!}{epsilon R^n} = frac{n!}{epsilon R^{n-k}} xrightarrow{R to infty} 0.$$
So we conclude from the Taylor expansion of $F$ that $$F(z) = sum_{n=0}^k frac{F^{(n)}(0)}{n!} z^n$$ is a polynomial of degree at most $k$. But as previously noted, $F$ has no zeroes. So $F equiv c$ for some $c inBbb C backslash{ 0}$! This means $g equiv c^{-1}$, and furthermore, $$widetilde{f}(z) = c^{-1}z^k forall z in Bbb C.$$
Supposing $k geq 2$, the polynomial $z^k - 1 in Bbb C[z]$ has at least two roots $xi_1, xi_2 in Bbb C$ which are certainly non-zero on account of the fact that $0^k neq 1$. The multiplicity of the root $xi_1$ is precisely $1$ because $$frac{d}{dz}Bigvert_{z=xi_1} z^k - 1 = k xi_1^{k-1} neq 0.$$ This allows us to conclude that $xi_1 neq xi_2$, and that $$widetilde{f}(xi_1) = c^{-1} xi_1^k = c^{-1} cdot 1 = c^{-1} xi_2^k = widetilde{f}(xi_2).$$ This result contradicts the fact that $widetilde{f}$ is injective. So $k=1$, and our final conclusion is that $$f(z) = widetilde{f}(z) + f(0) = c^{-1} z + f(0) forall z in Bbb C,$$ where for completeness's sake I recall that $c$ was necessarily non-zero.
answered Mar 3 '16 at 21:07
Open SeasonOpen Season
800518
$endgroup$
This proof uses only the Open Mapping Theorem and Cauchy's Estimates.
Suppose $f$ is entire and injective. Consider the injective entire function $$widetilde{f} colon z mapsto f(z) - f(0).$$ We see that $widetilde{f}(0) = 0$. Let $r > 0$, and consider the ball $B(0,r)$ of radius $r$ centered at the origin. By the Open Mapping Theorem, $widetilde{f}(B(0,r))$ is an open set containing the origin. By openness, there is some $epsilon > 0$ such that $B(0,epsilon) subset widetilde{f}(B(0,r))$. Since $widetilde{f}$ is injective, no point of $B(0,r)^c$ may be sent to any point of $B(0,epsilon)$. In other words, $$|z| geq r Rightarrow |widetilde{f}(z)| geq epsilon.$$
Now $widetilde{f}$ has a zero at the origin, so let's say this zero has order $k geq 1$. This means there is some entire function $g colon Bbb C rightarrow Bbb C$ such that $g(0) neq 0$ and $$widetilde{f}(z) = z^k g(z) forall z in Bbb C.$$
Since $widetilde{f}$ is injective, $widetilde{f}$ has no other zeroes apart from its zero at the origin. So $g$ has no zeroes. This means the function $F colon Bbb C rightarrow Bbb C$ defined by $$F(z) = frac{1}{g(z)} forall z in Bbb C$$ is entire and also has no zeroes. Write the Taylor expansion of $F$ at the origin as $$F(z) = sum_{n=0}^infty frac{F^{(n)}(0)}{n!}z^n.$$
Given any $R > r$, we can apply Cauchy's Estimates on the circle of radius $R$ to obtain bounds on the derivatives of $F$ at the origin. Indeed,
$$F^{(n)}(0) leq max_{|z| = R}|F(z)|frac{n! }{R^n}.$$ But we can apply the fact that $|widetilde{f}(z)| geq epsilon forall |z| geq r$ to notice that $$max_{|z| = R} |F(z)| = max_{|z| = R} frac{1}{|g(z)|} = max_{|z| = R} frac{|z|^k}{|widetilde{f}(z)|} leq frac{R^k}{epsilon}.$$ So if $n > k$, we have $$F^{(n)}(0) leq max_{|z| = R}|F(z)|frac{n!}{R^n} leq frac{R^k n!}{epsilon R^n} = frac{n!}{epsilon R^{n-k}} xrightarrow{R to infty} 0.$$
So we conclude from the Taylor expansion of $F$ that $$F(z) = sum_{n=0}^k frac{F^{(n)}(0)}{n!} z^n$$ is a polynomial of degree at most $k$. But as previously noted, $F$ has no zeroes. So $F equiv c$ for some $c inBbb C backslash{ 0}$! This means $g equiv c^{-1}$, and furthermore, $$widetilde{f}(z) = c^{-1}z^k forall z in Bbb C.$$
Supposing $k geq 2$, the polynomial $z^k - 1 in Bbb C[z]$ has at least two roots $xi_1, xi_2 in Bbb C$ which are certainly non-zero on account of the fact that $0^k neq 1$. The multiplicity of the root $xi_1$ is precisely $1$ because $$frac{d}{dz}Bigvert_{z=xi_1} z^k - 1 = k xi_1^{k-1} neq 0.$$ This allows us to conclude that $xi_1 neq xi_2$, and that $$widetilde{f}(xi_1) = c^{-1} xi_1^k = c^{-1} cdot 1 = c^{-1} xi_2^k = widetilde{f}(xi_2).$$ This result contradicts the fact that $widetilde{f}$ is injective. So $k=1$, and our final conclusion is that $$f(z) = widetilde{f}(z) + f(0) = c^{-1} z + f(0) forall z in Bbb C,$$ where for completeness's sake I recall that $c$ was necessarily non-zero.
answered Mar 3 '16 at 21:07
Open SeasonOpen Season
800518
answered Mar 3 '16 at 21:07
Open SeasonOpen Season
800518
answered Mar 3 '16 at 21:07
Open SeasonOpen Season
800518
answered Mar 3 '16 at 21:07
Open SeasonOpen Season
800518
800518
2
$begingroup$
Very nice! And you can even add that $c=1/(f(1)-f(0))$.
$endgroup$
– Unit
Mar 3 '16 at 21:15
add a comment |
2
$begingroup$
Very nice! And you can even add that $c=1/(f(1)-f(0))$.
$endgroup$
– Unit
Mar 3 '16 at 21:15
2
2
$begingroup$
Very nice! And you can even add that $c=1/(f(1)-f(0))$.
$endgroup$
– Unit
Mar 3 '16 at 21:15
$begingroup$
Very nice! And you can even add that $c=1/(f(1)-f(0))$.
$endgroup$
– Unit
Mar 3 '16 at 21:15
add a comment |
$begingroup$
Here is a (longer) proof using very little of complex analysis. Assume that $f:mathbb{C}tomathbb{C}$ is holomorphic and injective. The function $f$ extends to a holomorphic map of the Riemann sphere to itself, $mathbb{CP}^1tomathbb{CP}^1$. Indeed, choose any $z_0$ such that $f'(z_0)neq 0$ (if it doesn't exist then $f$ is constant (of course from injectivity we know that actually $f'neq0$ everywhere)). Then a small neighbourhood $Uni z_0$ is mapped bijectively to a small neighbourhood $Vni f(z_0)$. The function $1/(f(z)-f(z_0))$ is therefore bounded in $mathbb{C}-U$, hence by Riemann removable singularity theorem it extends to a holomorphic function on $mathbb{CP}^1-U$, therefore $f$ extends to a holomorphic map $mathbb{CP}^1tomathbb{CP}^1$.
As an application of Liouville's theorem, any holomorphic map $F:mathbb{CP}^1tomathbb{CP}^1$ such $F(z)neqinfty$ for $zneqinfty$ is a polynomial. If we wish, we can also avoid Liuoville theorem and use some topology.
If the order of pole of $f$ at $infty$ is $>1$ then $f$ is not injective in the neighbourhood of $infty$. Hence there is $ainmathbb{C}$ such that $f-az$ is holomorphic $mathbb{CP}^1tomathbb{C}$, hence it's bounded (being a map from a compact space), hence it's a constant: if $f-az$ is not constant then it is a map $mathbb{CP}^1tomathbb{CP}^1$ which is of positive degree but which is not surjective (as it avoids $infty$).
answered Mar 29 '11 at 20:21
user8268user8268
17k12947
$endgroup$
add a comment |
$begingroup$
Here is a (longer) proof using very little of complex analysis. Assume that $f:mathbb{C}tomathbb{C}$ is holomorphic and injective. The function $f$ extends to a holomorphic map of the Riemann sphere to itself, $mathbb{CP}^1tomathbb{CP}^1$. Indeed, choose any $z_0$ such that $f'(z_0)neq 0$ (if it doesn't exist then $f$ is constant (of course from injectivity we know that actually $f'neq0$ everywhere)). Then a small neighbourhood $Uni z_0$ is mapped bijectively to a small neighbourhood $Vni f(z_0)$. The function $1/(f(z)-f(z_0))$ is therefore bounded in $mathbb{C}-U$, hence by Riemann removable singularity theorem it extends to a holomorphic function on $mathbb{CP}^1-U$, therefore $f$ extends to a holomorphic map $mathbb{CP}^1tomathbb{CP}^1$.
As an application of Liouville's theorem, any holomorphic map $F:mathbb{CP}^1tomathbb{CP}^1$ such $F(z)neqinfty$ for $zneqinfty$ is a polynomial. If we wish, we can also avoid Liuoville theorem and use some topology.
If the order of pole of $f$ at $infty$ is $>1$ then $f$ is not injective in the neighbourhood of $infty$. Hence there is $ainmathbb{C}$ such that $f-az$ is holomorphic $mathbb{CP}^1tomathbb{C}$, hence it's bounded (being a map from a compact space), hence it's a constant: if $f-az$ is not constant then it is a map $mathbb{CP}^1tomathbb{CP}^1$ which is of positive degree but which is not surjective (as it avoids $infty$).
answered Mar 29 '11 at 20:21
user8268user8268
17k12947
$endgroup$
add a comment |
$begingroup$
Here is a (longer) proof using very little of complex analysis. Assume that $f:mathbb{C}tomathbb{C}$ is holomorphic and injective. The function $f$ extends to a holomorphic map of the Riemann sphere to itself, $mathbb{CP}^1tomathbb{CP}^1$. Indeed, choose any $z_0$ such that $f'(z_0)neq 0$ (if it doesn't exist then $f$ is constant (of course from injectivity we know that actually $f'neq0$ everywhere)). Then a small neighbourhood $Uni z_0$ is mapped bijectively to a small neighbourhood $Vni f(z_0)$. The function $1/(f(z)-f(z_0))$ is therefore bounded in $mathbb{C}-U$, hence by Riemann removable singularity theorem it extends to a holomorphic function on $mathbb{CP}^1-U$, therefore $f$ extends to a holomorphic map $mathbb{CP}^1tomathbb{CP}^1$.
As an application of Liouville's theorem, any holomorphic map $F:mathbb{CP}^1tomathbb{CP}^1$ such $F(z)neqinfty$ for $zneqinfty$ is a polynomial. If we wish, we can also avoid Liuoville theorem and use some topology.
If the order of pole of $f$ at $infty$ is $>1$ then $f$ is not injective in the neighbourhood of $infty$. Hence there is $ainmathbb{C}$ such that $f-az$ is holomorphic $mathbb{CP}^1tomathbb{C}$, hence it's bounded (being a map from a compact space), hence it's a constant: if $f-az$ is not constant then it is a map $mathbb{CP}^1tomathbb{CP}^1$ which is of positive degree but which is not surjective (as it avoids $infty$).
answered Mar 29 '11 at 20:21
user8268user8268
17k12947
$endgroup$
Here is a (longer) proof using very little of complex analysis. Assume that $f:mathbb{C}tomathbb{C}$ is holomorphic and injective. The function $f$ extends to a holomorphic map of the Riemann sphere to itself, $mathbb{CP}^1tomathbb{CP}^1$. Indeed, choose any $z_0$ such that $f'(z_0)neq 0$ (if it doesn't exist then $f$ is constant (of course from injectivity we know that actually $f'neq0$ everywhere)). Then a small neighbourhood $Uni z_0$ is mapped bijectively to a small neighbourhood $Vni f(z_0)$. The function $1/(f(z)-f(z_0))$ is therefore bounded in $mathbb{C}-U$, hence by Riemann removable singularity theorem it extends to a holomorphic function on $mathbb{CP}^1-U$, therefore $f$ extends to a holomorphic map $mathbb{CP}^1tomathbb{CP}^1$.
As an application of Liouville's theorem, any holomorphic map $F:mathbb{CP}^1tomathbb{CP}^1$ such $F(z)neqinfty$ for $zneqinfty$ is a polynomial. If we wish, we can also avoid Liuoville theorem and use some topology.
If the order of pole of $f$ at $infty$ is $>1$ then $f$ is not injective in the neighbourhood of $infty$. Hence there is $ainmathbb{C}$ such that $f-az$ is holomorphic $mathbb{CP}^1tomathbb{C}$, hence it's bounded (being a map from a compact space), hence it's a constant: if $f-az$ is not constant then it is a map $mathbb{CP}^1tomathbb{CP}^1$ which is of positive degree but which is not surjective (as it avoids $infty$).
answered Mar 29 '11 at 20:21
user8268user8268
17k12947
edited Mar 29 '11 at 20:31
answered Mar 29 '11 at 20:21
user8268user8268
17k12947
answered Mar 29 '11 at 20:21
user8268user8268
17k12947
answered Mar 29 '11 at 20:21
user8268user8268
17k12947
17k12947
add a comment |
add a comment |
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$begingroup$
en.wikipedia.org/wiki/Picard_theorem (which is overkill if this is homework or something)
$endgroup$
– yoyo
Mar 29 '11 at 17:06
2
$begingroup$
Maybe you can use the Weierstrass Factorization Theorem?
$endgroup$
– Adrián Barquero
Mar 29 '11 at 17:06
4
$begingroup$
By 1-1, do you mean injective or bijective?
$endgroup$
– lhf
Mar 29 '11 at 17:52
1
$begingroup$
@Arturo: you should let OPs do some of the fixing! It is a very instructive activity :)
$endgroup$
– Mariano Suárez-Álvarez
Mar 29 '11 at 19:56
1
$begingroup$
@Mariano: But it bugs me so...
$endgroup$
– Arturo Magidin
Mar 29 '11 at 20:40