# Blow-up. Graph of canonical map: is a complex manifold no closed or a topological space no closed

I am reading by myself this book http://tinyurl.com/37z4bbt. But to be honest, I have several problems to fully understand some part of the text. Maybe because I have not yet solid knowledge or I still need to learn more about the subject, hence asked a little help

To be exact, is this part of the construction concerning to Algebraic Blow-Up that I do not understand like trying...

I quote part of the building in which I do not understand how to deal with some concepts

Consider the canonical map from $\mathbb{C}^2$ to the projective line $\mathbb{CP}^1$ that associates with each point $(x,y)$ different from the origin, the line {$(tx, ty) : t \in \mathbb{C}$} passing through this point. The graph of this map is a complex 2-dimensional surface in the complex 3-dimensional manifold (the Cartesian product) $\mathbb{C}^2$ × $\mathbb{CP}^1$, which is not closed. To obtain the closure, one has to add the exceptional curve $E={0} \times \mathbb{CP^{1}} \subset \mathbb{C^{2}} \times \mathbb{CP^{1}}$.

then, I do not understand is how to handle this graph. If, as a topological space or as complex manifold. Well, in the sense of complex manifold, say it is not closed, it means that the manifold is not compact without boundary.

I am really a little confused

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It looks like it's not topologically closed since all the points in $\{0\}\times \mathbb{CP}^1$ are limit points of the surface but it does not contain them. –  Aaron Mazel-Gee Dec 1 '10 at 9:31

The book means the graph is not a closed subset of $\mathbb C^2 \times \mathbb CP^1$ (in the topological sense). Once you glue in $E$ (the exceptional curve or exceptional divisor), you obtain a closed set in $\mathbb C^2 \times \mathbb CP^1$. This will turn out to be a (complex) manifold, properly embedded in $\mathbb C^2 \times \mathbb CP^1$. The complement of $E$, by construction, will be biholomorphic to $\mathbb C^2$. (Note that this last fact means that you only need to check the manifold structure near the exceptional curve.)
This construction shows that you can take a complex 2 dimensional manifold, remove a point and then replace it by a copy of $\mathbb CP^1$ -- the local model is given by this construction. Let me know if you would like a bit more explanation of this.
If you forget the complex manifold structure, but instead just look at the manifold as a real manifold, you can interpret the construction as taking your complex manifold (e.g. $\mathbb C^2 = \mathbb R^4$ in this specific case) and making a connect sum with $\mathbb CP^2$ with its orientation reversed.