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Can anyone help me with a proof of the following claim (see for example the book Higher algebraic geometry of Olivier Debarre, proof of Proposition 1.43, page 31):

Let X be a complex manifold, and let W be a complex submanifold of X, with codimension $\geq 2$. Let $\pi :Y \rightarrow X$ be a bimeromorphic morphism, which is not an isomorphism, with the exceptional set $E$ so that $\pi (E)=W$ and $E$ is irreducible. Then there is a factorization $Y\rightarrow B_W(X)\rightarrow X$, where $B_W(X)$ is the blowup of $X$ at $W$.

There is also a statement for a universal property of blowup in Griffiths - Harris "Principles of algebraic geometry" but without proof as well. I also would like to know a proof of that fact.

Thank you very much.

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Is there a particular part where you are stuck? –  Yemon Choi Aug 23 '11 at 1:32
    
I understand the statements of those universal properties, and they seem plausible, but can not figure out how to prove them. By the way, the universal property of blowup in Griffiths-Harris that I mention is in the part "Blowup of manifold" in their book. –  anonymous Aug 23 '11 at 3:17
    
See the reference to Hartshorne mentioned by roy smith below and note that when Debarre applies the universal property of blowups he has reduced to the case that $Y$ is smooth. –  ulrich Aug 23 '11 at 6:45
    
Although I accepted the answer by Roy Smith, I would like to see whether there is an analytic proof of this universal property. One reason is to know that the map $Y\rightarrow B_W(Y)$ is holomorphic. Also, I have question of whether this universal property imply the universal property stated in Griffiths-Harris: If the fiber of any point on W is the projective space, then $Y=B_W(X)$. –  anonymous Aug 23 '11 at 13:23
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3 Answers

up vote 4 down vote accepted

I am kind of a rookie at this, but what if Y is a small resolution of a double point on a threefold X, with one dimensional excepTional locus. Then it seems false to expect a factorization through the blowup since the curve exceptional locus could not map onto the two dimensional exceptional locus of the blowup of X. what am i missing?

As pointed out by Anton, I am missing that the target space X is smooth. In that case the proof of Zariski's main theorem in Mumford Cx Proj Vars, p.49 shows there the exceptional locus E contains a cartier divisor through every point. Hence when E is irreducible it is cartier. Then the universal property of blowing up in Hartshorne implies the factorization exists in the algebraic category. The proof probably also works in the analytic category.

As to an analytic argument for the G-H universal property, it seems the hypothesis there is that you have a holomorphic map of manifolds Y-->X, the inverse image E of a certain submanifold W is also a submanifold of codimension one, and the fiber over every point of W is a projective space of dimension equal to the difference in the dimensions of the two submanifolds. Is that right?

Then you want a factorization through the blowup of X along W and you want it to be an isomorphism. Assume for simplicity the manifolds are compact. Then think what a blow up means. You are replacing the target submanifold W by its projectivized normal bundle in X. Hence the natural factorization would be via the derivative of the original map f. In fact an examination of Hartshorne's factorization will show it is simply the derivative.

Your hypotheses imply that at each point of E, the kernel of the derivative equals the tangent space to the fiber. Hence the tangent space to E surjects onto the tangent space space to W, and the image of the full tangent space of Y is of dimension one larger than that of W. Thus the derivative defines an injection from the normal line bundle of E, into the normal bundle of W in X. That is precisely an induced map from E to the exceptional locus of the blowup of X along W.

This map is holomorphic on E, since it is the derivative of a holomorphic map. One needs only check that it glues in as a continuous, hence holomorphic, extension of f, and this needs be done only in the normal direction to E, where it is essentially the definition of a derivative as a limit of difference quotients.

To see that the factorization is an isomorphism, it suffices to check it is bijective, which need only be checked on E. There we have on each fiber of f, a surjective holomorphic map of projective spaces of the same dimension. By the dimension counts above a regular value of this map is also regular for the factorization, hence each holomorphic map of projective spaces also has degree one, hence is an isomorphism.

Here is the universal property in a nutshell:

1) Blowing up an ideal is a functor.

I.e. if f:Y-->X is a map, and (g1,...,gr) is a ideal of functions on X, then f lifts to a morphism from the blowup of Y along the ideal (g1of,...,.grof), to the blowup of X along (g1,...,gr).

2) Blowing up a principal ideal does nothing.

Hence, if the pull back of an ideal is a principal ideal, i.e.defines a cartier divisor, then the original map factors through the blowup of the target space.

This universal property is essentially trivial. I.e. if you get away from all the proj's and gr's the blowup of the subvariety of X defined by {g0,...,gn} is just the closure in XxP^n of the graph of the meromorphic function g:X-->P^n, defined by the {gj}.

hence if f:Y-->X is holomorphic then so is (fx1):YxP^n-->XxP^n, and it takes the closure of the graph of (gof) into the closure of the graph of g. Moreover if n = 0, nothing happens. Done. [The gr, proj stuff comes in to show this is all independent of choice of generators of the ideals.]

the reference below to Fischer seems excellent. the access i have through Amazon only gives the special case of a one point blowup, but by implication, that case is crucial. We can see this is true by observing that our definition of the local blowup agrees with the pull back by the map g, of the blowup of the point 0 in C^(n+1).

I.e. if we blowup the point 0 on C^n+1, by taking the closure in C^(n+1)xP^n of the graph of the map defined by the coordinate functions on C^(n+1), and then map X into C^n+1 by the map g, the induced map of XxP^(n) into C^(n+1)xP^n pulls back the blowup of 0 in C^(n+1) to the blowup of the zero scheme of g in X.

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I was going to say the same thing, but figured "complex manifold" probably means that it has to be smooth. Is the exceptional locus over a smooth point always a divisor? –  Anton Geraschenko Aug 23 '11 at 3:51
    
ok let me be more careful. from page 104 of the 1st ed. of shafarevich and page 48 of mumford's cx proj varieties, it seems the inverse image of W is of codimension one. from page 164 of hartshorne it seems the factorization exists iff the pullback ideal of W to Y is an invertible sheaf of ideals. now it seems that the sheaf of ideals of an irreducible subvariety of codimension one is invertible on a smooth variety, so I need still to know that Y is smooth along the pullback of W. ??? –  roy smith Aug 23 '11 at 4:27
    
In the proof of Debarre mentioned in the question, $Y$ is smooth when the universal property of blowups is used, so I think what you say is correct. –  ulrich Aug 23 '11 at 6:45
    
Thank you. I forgot to say that Y is smooth. The reference in Hartshorne solves my question. However, do you have an analytic proof of this fact? Also, does this universal property implies the following universal property in Griffiths-Harris: Let X, Y, E, W and $\pi$ be are as above, and the fiber at any point on W is the projective space. Then Y is the blowup $B_W(X)$. –  anonymous Aug 23 '11 at 13:01
    
By analytic method, I mean some kind of Hartogs principle, which I guess can be used to prove the result. @Roy Smith: Under the conditions in my question, then in the book of Debarre it is shown that E is a hypersurface. –  anonymous Aug 23 '11 at 13:07
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The book by Fischer, "Complex analytic geometry" has a nice treatment of the blow-up and its universal property in chapter 4. The proofs are given in the analytic category.

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Thank you Sylvain. It seems interesting. I will look at the book. Do you know is there an electric file of that book? The book in my school's library has been checked out. –  anonymous Aug 25 '11 at 3:10
    
You can email me (you'll find my address on math.utoronto.ca) –  Sylvain Bonnot Aug 25 '11 at 5:18
    
Thank you for your offer. I asked for interlibrary loan, and am waiting for the hardcopy. –  anonymous Aug 26 '11 at 12:49
    
I got the book. It is very good. It is not very thick, it is balance between not going very detail into proofs but it gives a lot of helpful examples. It is kind of helps people strengthen understanding after having some knowledge in the subject. –  anonymous Sep 2 '11 at 3:45
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Thank you Roy Smith for your excellent answer.

So as I understand, in the smooth case, the universal property goes as follows:

Let $f:Y\rightarrow X$ be a surjective holomorphic map between complex manifold, let W be a submanifold of X so that its inverse image is a submanifold E of Y. Assume moreover that both E and W are irreducible. We can work locally, so can assume that both E and W have good tubular neighborhoods NE and NW, which are isomorphic to their normal bundles. Now the derivative of f will give a lifting map

$F: B_EY\rightarrow B_WX$. In case $E$ is a hypersurface then $B_EY=Y$ and we obtain the universal property referred to by Debarre.

Now for the universal property in Griffiths-Harris:

Now if f is moreover biholomrphic from $Y-E$ to $X-W$ then the map $F$ must be surjective? (If instead we just ask that $f$ is finite to one on $Y-E$, do we still have this property? And in general, if we just ask $f$ to be surjective, is the map F surjective?)

Now if moreover E is a hypersurface, then F maps each fiber on E (which is a $P^k$) to each fiber of the exceptional divisor of the blowup $B_WX$ (which is also a $P^k$). This map $F$ restricts to $P^k$ is holomorphic surjective to itself, and thus must be finite- to-one (since $P^k$ is Kahler). Thus the map $F$ restricted to $E$ is finite-to-one. Then since for points in a neighborhood of $E$ but not on $E$, the degree of $F$ is $1$, it must be so on $E$ as well; hence an isomorphism.

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surjectivity follows from properness. –  roy smith Aug 25 '11 at 13:33
    
i.e. assume the original map is proper, as occurs for compact manifolds, or any restriction of a map of compact manifolds over any subset of the target. then the induced map of blowups is also proper, hence closed. Thus if the original map was surjective, the induced map is dense and closed, hence surjective. –  roy smith Aug 26 '11 at 2:33
    
Thanks. I see. In the case under the conditions of G-H universal property, the properness of the map f is easy to check, or am I wrong? –  anonymous Aug 26 '11 at 12:51
    
i have not checked their hypotheses, and that is why i assumed compactness to make properness automatic, but properness is usually assumed. –  roy smith Aug 28 '11 at 3:00
    
look at mumford's lemma 3.11 page 44 of his cx alg geom 1, and see if you can crank up the proof to a proof that a map with compact inverse image of a compact set is proper on some nbhd, but i am not sure it is true. –  roy smith Aug 28 '11 at 3:04
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