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. 
 A: 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
A: 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).
A: 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.
