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Recall that give a map $f : (X,\mathcal{O}_X) \to (Y,\mathcal{O}_Y)$ of ringed spaces and a sheaf $\mathcal{F}$ on $Y$ we can form the pullback $f^\ast \mathcal{F} := f^{-1}\mathcal{F}\otimes_{f^{-1}\mathcal{O}_Y} \mathcal{O}_X$.

Now the pullback of sheaves appears in both chapters 2.5 and 2.7 of Hartshorne. In particular, it is used to define a correspodence between line bundles $\mathcal{L}$ on a scheme $(X \to \operatorname{Spec} A)$ and maps to projective space $\Bbb{P}^n_A$. More explicitly, for any morphism $\varphi : X\to \Bbb{P}^n_A$ we have $\varphi^\ast \mathcal{O}(1)$ being a line bundle generated by the linear polynomials $x_0,\ldots,x_n$. Conversely given $n+1$ global sections $s_0,\ldots,s_n$ of a line bundle $\mathcal{L}$ on $X$, we have a unique morphism $\varphi : X \to \Bbb{P}^n_A$ such that $\varphi^\ast(x_i) = s_i$.

Problem 1: My first problem I have with $\varphi^\ast$ is say I have the $x_i$'s as above. What are the "canonical induced sections" $s_i \in \varphi^\ast\mathcal{O}(1)$? All I can say now is that each $x_i$ gives a map from $\mathcal{O}_{\Bbb{P}^n_A} \to \mathcal{O}(1)$ but how does this give us our $s_i$? Is there a more concrete description of what these actually are?

Problem 3: Can we see explicitly "in coordinates" what the map $\varphi : X\to \Bbb{P}^n_A$ determined by these sections is?

Problem 2: My second problem with the definition of this $\varphi^\ast$ is getting hold of its global sections. From the definition I gave in the beginning of the pullback, it seems that definition is not much help in practical situations! Say I want to compute global sections of $\varphi^\ast \mathcal{O}(1)$ where $\varphi$ is the closed immersion of the conic $ V(xy - z^2)$ in $\Bbb{P}^2$. How can I do this? What about computing global sections of the bullback bundle by say $\varphi: \Bbb{P}^n\setminus V(x_0,x_1,x_2) \to \Bbb{P}^2$?

Last and finally, is there a way to think of this pullback operation that can significantly enhance my feel for it? Let me give an example of what I mean by this. When I first learned about the tensor product of sheaves I was very fearful because I didn't know how the sheaf behaved (for example what on earth are the global sections?). However if we tensor two quasi-coherent sheaves, then on an affine I know exactly what the tensor product is: it behaves just like a module!

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  • $\begingroup$ The details of this bijection are given very explicitly in (EGA, II, 4.2.3). $\endgroup$ – user314 Oct 25 '13 at 12:51
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    $\begingroup$ For Problem 1, one concrete way to understand the pullback might be via rational functions. There is a non-canonical way to identify sections of a bundle with rational functions having certain prescribed poles (explained e.g. in Volume 2 of Shafarevich's book). Given this, you can pull back sections of a bundle just by the usual pullback of rational functions. Of course this has certain drawbacks --- it's not canonical, and when $\varphi$ isn't dominant you need to worry about indeterminacy. Still, I think it helps to give some insight. $\endgroup$ – user64687 Oct 25 '13 at 14:18
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    $\begingroup$ Forget about the explicit construction of the pullback, it isn't useful at all. Pullback = left adjoint to direct image. Locally the pullback just tensors with the corresponding ring extension. This is how one works with pullbacks. P1: What do you mean by concrete? In what terms do you want to describe them? We have sections $x_i : \mathcal{O} \to \mathcal{O}(1)$, so we can pull them back to $\phi^* x_i : \mathcal{O} \to \phi^* \mathcal{O}(1)$. So this is just functoriality. $\endgroup$ – Martin Brandenburg Oct 26 '13 at 8:57
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    $\begingroup$ P3: What do you mean by coordinates? Perhaps $k$-valued points? Then the map is given by $x \mapsto [s_0(x):\dotsc:s_n(x)]$. By this one actually means $[s_0(x)/s_i(x) : \dotsc : s_n(x)/s_i(x)]$ when $s_i(x)$ generates $\mathcal{L}$, and $s_j(x)/s_i(x)$ is the unique element of $k$ with $s_j(x)/s_i(x) \cdot s_i(x) = s_j(x)$ in $\mathcal{L}$. $\endgroup$ – Martin Brandenburg Oct 26 '13 at 9:00
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    $\begingroup$ An off topic note: In my opinion, it isn't appropriate to use as a pseudonym the name and image of an actual living mathematician. $\endgroup$ – David E Speyer Oct 30 '13 at 21:02
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Pullback of invertible sheaves corresponds to pull-back of line bundles in the geometric sense.

In usual topology, a (complex) line bundle is a family of one-dimensional $\mathbb C$-vector spaces varying over a topological space $X$. More precisely, it is a topological space $V$, equipped with a continuous map $V \to X$, a zero section $X \to V$, a scalar multiplication action $\mathbb C \times V \to V$ compatible with the projections to $X$, an addition map $V \times_X V \to V$, again compatible with the projections to $X$, all satisfying some evident axioms, which more or less amount to requiring that these structure endow each fibre $V_x$ of $V$ over a point $x$ of $X$ with the structure of a one-dimensional $\mathbb C$-vector space, in such a way that $V$ is locally trivial.

Now if $\varphi: Y \to X$ is a continuous map, we may form the fibre product $\varphi^*V := V\times_X Y,$ and this is a line bundle over $Y$.

If you just think about how fibre product is defined, you see that the fibre of $\varphi^*V$ over a point $y \in Y$ is canonically identified with the fibre of $V$ over $\varphi(y)$. So if we think of $V$ as being the family $V_x$ of lines parameterized by the points $x \in X$, then $\varphi^*V$ is the family of lines $V_{\phi(y)}$ parameterized by $y \in Y$.

We may form the sheaf of sections $\mathcal L$ of $V$; this is a locally free sheaf of rank one over the sheaf of continuous $\mathbb C$-valued functions on $X$. The sheaf of sections of $\varphi^*V$ is then $\varphi^*\mathcal L$.

There is an exercise somewhere in Hartshorne which describes the analogue of the above theory of vector bundles in the algebro-geometric setting; everything goes over more-or-less the same way. I am now going to work in the algebro-geometric setting.


It is not so easy to relate the sections of $V$ to those of $\varphi^*V$; it depends very much on the space $Y$ and the morphism $\varphi$. For example, if $Y$ is just a point $x \in X$, then $\varphi^*V$ is just the line $V_x$, and its space of sections is one-dimensional (indepdently of what $V$ is). But it might be that $V$ has no sections that are non-zero at $x$.

More generally, a non-trivial bundle on $X$ might pull-back to something trivial on $Y$.

So you have to analyze this question on a case-by-case basis. Cohomology is one of the basic tools available here, but I am guessing that you're not at the point of using cohomology yet.

E.g. if $Y = \mathbb P^1$ and $\varphi: Y \to \mathbb P^2$ embeds $Y$ as a plane conic, then this image is a degree two curve, so $\mathcal O(1)$ pulls back to a degree two sheaf on $Y$, i.e. to $\mathcal O(2)$. Thus its space of global sections is three dimensional. So in this case $\varphi^*$ induces an isomorphism on spaces of global sections.

E.g. If $Y = \mathbb P^1$ but $\varphi: Y \to \mathbb P^2$ embeds it as a line, then $\varphi^*\mathcal O(1)$ is just $\mathcal O(1)$ on $\mathbb P^1$ (a line is a degree one curve), and so the map on sections is surjective, with a one-dimensional kernel (which is precisely the linear form cutting out the image of $\varphi$).


As Martin Brandenburg indicates in an answer, the sections $s_i$ give a projective embedding in the following concrete way: locally we may trivialize $V$, and so regard the $s_i$ as simply functions; we thus obtain a morphism $x \mapsto [s_0(x): \cdots : s_n(x)].$ Note that if we change trivialization, then the interpretation of the $s_i$ as functions changes by a nowhere zero function (the same function for all the $s_i$), which means that the point in $\mathbb P^n$ doesn't change. Thus the map is well-defined independent of the trivialization.


Finally, it might be helpful to know that in topology, the complex line bundles are classified (up to isomorphism) by homotopy classes of maps to $\mathbb C P^{\infty}$. Again, the map is given by pulling back $\mathcal O(1)$.

The situation in algebraic geometry is analogous to this, but more rigid: topological isomorphism is much less rigid than algebraic isomorphism (hence only the homotopy class of $\varphi$ matters), and all line bundles have lots of sections. In algebraic geometry, only sufficiently positive (i.e. sufficiently ample) bundles arise from maps to projective space.

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