Universal property of blowups 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. 
 A: 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. 
A: 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.
A: 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.   
