Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose that $(X,\mathcal{O}_X),(Y,\mathcal{O}_Y)$ are two ringed spaces and $(f,f^{\sharp}):(X,\mathcal{O}_X)\longrightarrow(Y,\mathcal{O}_Y)$ is a morphism of these two ringed spaces.I wonder why we define $f^\sharp$ going from $\mathcal{O}_Y$ to $f_* \mathcal{O}_X$ instead of from $f^{-1}\mathcal{O}_Y$ to $\mathcal{O}_X$.Though I feel this question is a bit metaphysical,I still want to learn some explanation about it.Are there some advantages in the first definition or just due to historical reasons?And I would appreciate it if someone would like to give me some hints on it.

share|improve this question
2  
The definition of the inverse image sheaf involves a direct limit. This, in my mind, makes it less basic. –  Zhen Lin Aug 10 '11 at 15:10
    
Thank you very much.I got it. –  user14242 Aug 10 '11 at 15:13
    
@Dylan: could you post that as an answer? –  Sean Tilson Aug 10 '11 at 16:40
4  
A "symmetric" way to think of it as follows: whenever $f(U) \subset V$ (for $U \subset X, V \subset Y$ open), there is a ring-homomorphism from $\Gamma(V, \mathcal{O}_Y) \to \Gamma(U, \mathcal{O}_X)$ (and these are compatible). –  Akhil Mathew Aug 10 '11 at 16:52
1  
As Zhen says, the inverse image $f^{-1}\mathscr{O}_Y$ requires you to take direct limits, and moreover you still have to sheafify. So, at least in my opinion, it's a lot harder to think about than $f_*\mathscr{O}_X$. I think in a lot of instances, one uses $f^{-1}\mathscr{O}_Y$ via the adjunction with $f_*$ (instead of dealing directly with the definition of $f^{-1}\mathscr{O}_Y$). –  Keenan Kidwell Jul 18 '12 at 22:16

1 Answer 1

up vote 6 down vote accepted

As $f_*$ is right adjoint to $f^{-1}$ (see Wikipedia; this is also Ch. 2, Ex. 1.18 in Hartshorne), by picking one of these definitions you aren't really missing anything. The standard definition feels right to me because this is how smooth functions on a manifold behave: if $f\colon M \to N$ is a smooth map of manifolds and $V \subset N$ is open, then even without the notion of a sheaf it is natural to pull back smooth functions $V \to \mathbf R$ to smooth functions $f^{-1}(V) \to \mathbf R$.

Vakil's notes usually present some motivation for their definitions. The book by Eisenbud and Harris is also good about this.

Added much later. The relevant section in the Stacks project introduces the notion of an "$f$-map of sheaves" in Definition 21.7. I think this is conceptually a lot easier to swallow, particularly when it comes to composing morphisms of ringed spaces.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.