# Monomorphisms of sheaves gives an injection of stalks

How do I show that a monomorphism $F \rightarrow G$ of sheaves induces an injection on stalks?

When showing that monomorphism is an injection on sets one uses the maps $x \mapsto a$ and $x \mapsto b$ where $a,b$ are some fix elements such that they have the same image under the monomorphism. Then by using the monomorphism property we must have $a = b$. But since we are dealing with sheaves we are supposed to construct a skyscraper sheaf and two maps that map "almost all" section of the skyscraper sheaf to $a$ and $b$ (respectively), or at least elements that are "connected" to $a$ and $b$ (since these are representations of elements in the stalk there are other options depending on which open set we are taking). When I say "almost all" and "connected" it is just my visualization of how it should be.

I'm not able to neither define the correct skyscraper sheaf nor the maps to $F$ (especially such that the maps commute with the restriction maps). I was hoping for this being true which might have helped me but now I am stuck and need some help with this problem.

Edit: I am still interested in how the problem will be solved by using a skyscraper sheaf.

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In fact, more is true. If $U$ is an open set, then $F(U) \to G(U)$ is a monomorphism for $F \to G$ a monomorphism of sheaves. To see this, use the constant sheaf $\mathbb{Z}_U$ over $U$, and its extension by zero $j_!(\mathbb{Z}_U)$, which is a sheaf on $X$ (where $j: U \to X$ is the inclusion). To hom from $j_!(\mathbb{Z}_U)$ into a sheaf $H$ (in the category of sheaves over $X$) is the same thing as giving a section of $H$ over $U$, by the universal property of extension by zero (for the lower shriek $j_!$ is left adjoint to the restriction to $U$, so $\hom_X(j_!(\mathbb{Z}_U, H) = \hom_U(\mathbb{Z}_U, H|_U) = \Gamma(U, H)$). As a result, by definition of a monomorphism, $F \to G$ monomorphic implies that $\Gamma(U, F) \to \Gamma(U, G)$ is injective for each $U$ open.

This implies injectivity on the stalks because filtered colimits are exact in the category of abelian groups (or modules over a ring).

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Thank you. I do not know what it means for a colimit to be filtered but I will see if I manage to solve the problem (given your answer) tomorrow. @All: I am still interested in how the problem will be solved by using a skyscraper sheaf. – erwin Nov 14 '10 at 0:46
Dear Akhil, Just to be precise (assuming I'm not confusing myself): do you mean $j_! \mathbb Z_U$, where $j: U \to X$ is the inclusion? Regards, – Matt E Aug 11 '11 at 0:03
@Matt: Dear Matt, thanks for the correction. – Akhil Mathew Aug 11 '11 at 0:07

There is a nice proof of this on page 7 of Kiran Kedlaya's notes (and also that surjectivity carries over to stalks) http://ocw.mit.edu/courses/mathematics/18-726-algebraic-geometry-spring-2009/lecture-notes/MIT18_726s09_lec03_sheaves.pdf

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Thank you for the link but I don't think it solves my problem. For example, how do I know that a monomorphism f of sheaves (meaning that if fg = fh then g => h) implies that f(U) is injective for each U (this is actually the kernel of my issue)? If I understand this step I will be able to prove that we have an injection on stalks. But using Akhils answer + your link simply does the work. – erwin Nov 14 '10 at 0:41

This is one way to see it. The stalk functor (that sends each sheaf $F$ to its stalk $F_x$ at fixed point $x$) is a filtered colimit, and they are known to commute with finite limits.

Morphism $f$ is monomorphism iff diagram $\newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} \bullet & \ra{1} & \bullet & \\ \da{1} & & \da{f} \\ \bullet & \ra{f} & \bullet \\ \end{array}$ is pullback, which is finite limit.

So, stalk functor preserves pullbacks, and therefore monomorphisms.

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