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There is a point about maps betwen line bundles (continuous or smooth or holomorphic --- I don't think it matters for this question) that is glossed over in many texts.

A map from one line bundle $N_1$ to another $N_2$ over the same space $X$ is a map $h$ that makes a commutative triangle with the projections to $X$.

These maps form a vector space over the base field, and we will call this vector space $\mbox{Hom}_X(N_1,N_2)$. Let's fix the base field to be $\mathbb C$ and take the fibres of the $N_i$ to be copies of $\mathbb C$, to make things concrete.

On the other hand, maps between $N_1$ and $N_2$ can also be thought of as (continuous, smooth, holomorphic) global sections of the line bundle $N_1^*\otimes N_2$, i.e. as elements of $H^0(X;\mathcal O(N_1^*\otimes N_2))$, where $\mathcal O$ denotes the sheaf of continuous or smooth or holomorpic sections, as appropriate.

Question:

Why is $\mbox{Hom}_X(N_1,N_2)=H^0(X;\mathcal O(N_1^*\otimes N_2))$, at least as sets? (Once set-theoretic equivalence is established, the vector space isomorphism should come for free.)

The reason why these two view points on line bundle maps are equivalent is not explained in any of the standard texts that I use.

I understand the correspondence at the level of a single fibre, i.e. the linear algebra of why maps from one copy of $\mathbb C$ into another copy are given by $\mathbb C^*\otimes\mathbb C$. But thinking at the level of open sets or globally, a bundle map from the first view point is a map between the total spaces of $N_1$ and $N_2$ that commutes with projections to $X$. From the second point of view, it is a map from the space $X$ into the total space of $N_1^*\otimes N_2$ that commutes with a projection to $X$. How can I reconcile these?

(As part of the question, I'm also not sure how the projection of $N_1^*\otimes N_2$ to $X$ is determined by the data of $N_1$ and $N_2$. I understand how the transition functions of the tensor product relate to those of $N_1$ and $N_2$ --- I'm not asking about that. How is the projection map for the tensor product is obtained?)

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Hint. Think first about the case where X is a point, where this reduces to linear algebra. The general case is exactly the same.

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  • $\begingroup$ As for your parenthetical question as the end: you should certainly review the construction of the tensor product of bundles —indeed, it is rather surprising that you are even asking your main question if you are quite unsure about what the tensor product of bundles is! $\endgroup$ Commented Jan 1, 2015 at 5:31
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    $\begingroup$ Well, i did. Exactly the same thing happens over a point and over a general X. A map f of bundles $M\to N$is a choice, for each point x of X, of a linear map $f_x:M_x\to N_x$ over the corresponding fibers, which is, by linear algebra, the same thing as an element $t_x$ of $M_x^*\otimes N_x$, which is the fiber of the bundle $M^*\otimes N$ over x (essentially by definition) This defines a section t of the bundle $M^*\otimes N$. $\endgroup$ Commented Jan 1, 2015 at 5:43
  • $\begingroup$ That is what I suggested you to think about in the case of a point—notice that the assignment from f to t does not depend at all on whether X is a point or not, and that was my point. Of course, for this to work one has to prove that the section is smooth if f is a smooth bundle map and that one does obtain a bijection in this way, and so on. To do that, you will need to review the construction of the tensor product, as among other things you will need to know its projection map is. $\endgroup$ Commented Jan 1, 2015 at 5:47
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    $\begingroup$ I'd rarher not. I am a firm believer thar understanding is a process, and often comments capture that process much better than an answer which appears ex nihilo. $\endgroup$ Commented Jan 1, 2015 at 6:09
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    $\begingroup$ Dear MathsByTheSea, yes there is a canonical isomorphism of complex vector spaces $Hom(M,N)=\Gamma(X,\mathcal O(M^*\otimes N)) $ valid for vector bundles $M, N$ of any rank. And, yes, the global secctions of any vector bundle $V$ are indifferently denoted $\Gamma(X,V)$ or $H^0(X,V)$. $\endgroup$ Commented Jan 2, 2015 at 20:31

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