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I've recently started reading about sheafs and ringed spaces (at the moment, primarily on wikipedia). Assuming I'm correctly understanding the definitions of the direct image functor and of morphisms of ringed spaces, a morphism from a ringed space $(X, O_X)$ to a ringed space $(Y, O_Y)$ is a continuous map $f\colon X\to Y$ along with a natural transformation $\varphi$ from $O_Y$ to $f_*O_X$.

Why does the definition require $\varphi$ to go from $O_Y$ to $f_*O_X$ as opposed to from $f_*O_X$ to $O_Y$?

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Why do you expect/want the natural transformation to go the other way? – Nils Matthes Nov 7 '12 at 21:53
@NilsMatthes: for most of the first algebraic/topological/etc structures a typical student meets, a morphism from $X$ to $Y$ consists of a function, or a few functions, each going from something-to-do-with-$X$ to something-to-do-with-$Y$. The first encounter with a definition where one of the functions goes the other way is (for most people, I think) a bit disorienting — and ringed spaces may well be the first such case that a student meets. – Peter LeFanu Lumsdaine Nov 8 '12 at 6:52
up vote 13 down vote accepted

1) The model one has to keep in mind is that $X$ and $Y$ are geometric spaces equipped with functions of a certain degree of regularity defined on their open subsets.
A typical example would be that $X$ and $Y$ are $C^\infty$ manifolds, that for $V\subset Y$ one takes $\mathcal O_Y(V)=C^\infty (V)$ and similarly for $X$.
If $f:X\to Y$ is a $C^\infty$ map one gets for each open subset $V\subset Y$ a composition map $f^{*}_V : C^\infty (V)\to C^\infty (f^{-1}V):g\mapsto g\circ f$ and letting $V$ vary the $f^{*}_V$ yield a morphism of sheaves of rings on $Y$:$$\mathcal O_Y\to f_*\mathcal O_X$$ Do you see that there is no way the arrow could go in the other direction?

2) For general ringed spaces the $\mathcal O_Y(V) $ 's are no longer required to be functions but are abstract rings and the $f^{*}_V : \mathcal O_Y (V)\to \mathcal O_X (f^{-1}V)$ are no longer compositions but are given as part of the structure: they are exactly the datum of a morphism of sheaves $\mathcal O_Y\to f_*\mathcal O_X$ .

Beware that a morphism of ringed spaces is a very general concept :
Any morphism of rings $A\to B$ can be seen as a morphism of one-point ringed spaces $(\lbrace * \rbrace ,B)\to (\lbrace * \rbrace,A)$ .
But usually one only studies locally ringed spaces, a non-full subcategory of the category of all ringed spaces, the best known example being the category of schemes.

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Think about what it means to give a morphism from $\mathcal O_Y$ to $f_* \mathcal O_X$: it means that for every open set $V \subset Y$, there is a map $$\mathcal O_Y(V) \to \mathcal O_X\bigl( f^{-1}(V) \bigr).$$

If you imagine that $\mathcal O_X$ and $\mathcal O_Y$ are supposed to be some sorts of "sheaves of functions" on $X$ and $Y$, then this accords perfectly with the intuition that a morphism of ringed spaces should allow us to "pull back" functions.

Indeed, in concrete examples (such as smooth manifolds equipped with the structure sheaf of smooth functions), the map $\mathcal O_Y \to f_* \mathcal O_X$ is just the pull-back map on functions.

A morphism in the opposite direction doesn't have any analogous intuitive interpretation, and doesn't accord with what happens in the key motivating examples.

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The data of $\phi$ is a bunch of maps $O_Y(U) \to (f_\ast O_X)(U) = O_X(f^{-1}(U))$. If you're thinking of $O_Y$ as being a sheaf of functions of some sort, you can think of $\phi$ as giving directions for pulling back functions on $Y$ to functions on $X$ (or subsets thereof). Going the other way wouldn't make sense.

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A ringed space is a pair $(X, \mathcal{O}_{X})$, where $X$ is a topological space and $\mathcal{O}_{X}$ is a sheaf of rings on $X$.

Then a morphism of ringed spaces from $(X,\mathcal{O}_{x})$ to $(Y,\mathcal{O}_{Y})$ is a pair $(f, \phi)$ where $f \colon X \to Y$ is a morphism of topological spaces and $\phi \colon \mathcal{O}_{Y} \to f_{*}\mathcal{O}_{X}$ is a morphism of sheaves (of rings on $X$).

Consider a module $\mathcal{F} \in \textrm{Mod}(\mathcal{O}_{X})$, then $f_{*}(\mathcal{F})$ is a sheaf on $Y$ and an $f_{*}\mathcal{O}_{X}$-module (which is also an $\mathcal{O}_{Y}$-module), so the push forward in this case is a functor from $\mathcal{O}_{X}$-modules to $\mathcal{O}_{Y}$ modules.

This is the direct image functor, which may be more conveniently written as a functor of abelian categories: $$f_{*} \, \colon \, \textrm{Mod}(\mathcal{O}_{X}) \to \textrm{Mod}(\mathcal{O}_{Y})$$

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Why is $f_\ast (\mathcal F)$ also an $\mathcal O_Y$-module? – Exterior Nov 7 '15 at 8:57

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