Say i am given the sheaf:
\begin{eqnarray*} M_X: {{Op}_X}^{op} & \longrightarrow & \text{Mod}(k)\\ U & \longmapsto & M_X(U) := \lbrace f: U \longrightarrow M \mid \forall_{x \in U }\exists_{V \in I_x}:f\mid_V = \text{const} \rbrace \end{eqnarray*}
which is called the constant sheaf with values in the category of modules.
I need to show the following:
Every stalk of $M_X$ is isomorphic to $M$:
Can someone show me the proof in terms of inj/surj ?
Proof:
Let $U$ be a nbh of $x \in X$ and \begin{eqnarray*} \phi_{U,x} : M_X(U) & \longrightarrow & M\\ f & \longmapsto & \phi_{U,x}(f) := f(x) \end{eqnarray*} be the evaluating map (k-module homomorphism). We have to show, that the evaluating map \begin{eqnarray*} (M_X)_x & \longrightarrow & M\\ f_x & \longmapsto & f(x) \end{eqnarray*} is an isomorphism.
Inj.: Let $f(x) = 0$. It ex. a nbh $V$ of $x$ such that $f\mid_V = 0$. because $f$ is locally constant. That induces that the germ is 0.
Surj.: For $m \in M$ choose $f_x$ the constant section \begin{eqnarray*} f: X & \longrightarrow &M\\ p & \longmapsto & f(p) := m \end{eqnarray*} for all $p \in X$. So obv the $f_x$ is $m$.