Question: 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](https://math.stackexchange.com/questions/10047/can-every-element-in-the-stalk-be-represented-by-a-section-in-the-top-space) 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.

Answer: 
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>
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.