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: 
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).
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.