TITLE: Fibre of a sheaf of $\mathcal{O}_X$-modules.
QUESTION [9 upvotes]: Let $\mathcal{F}$ be a locally free sheaf of modules of rank $n$ over some complex manifold $X$. 
To $\mathcal{F}$ we can associate a vector bundle call, $\pi: V \to X$. 
The fiber of the sheaf $\mathcal{F}$ is $\mathcal{F}(x) := \mathcal{F}_x \otimes_{\mathcal{O}_x} k(x)$ where $k(x)$ is the residue field for some point $x \in X$. Note that this is definitely a vector space over $k(x)$. 
Is $\mathcal{F}(x) \cong \pi^{-1}(x)$? 
It seems like a basic question but I cannot find a straightforward answer.
Using the correspondence of between locally free sheaves of modules and vector bundles $\pi^{-1}(x) = \mathcal{F}_x / m_x \mathcal{F}_x$. 
We have the following:
$$\mathcal{F}_x / m_x \mathcal{F}_x \cong \mathcal{F}_x \otimes_{\mathcal{O}_x} \mathcal{O}_x/m_x' \mathcal{O}_x \cong \mathcal{F}_x / m'_x \mathcal{F}_x$$. 
Then this boils down to show that the maximal ideal's, $m'_x \subset \mathcal{O}_x$, acts on $\mathcal{F}_x$ as $m'_x \mathcal{F}_x= m_x \mathcal{F}_x$

REPLY [4 votes]: Yes, this is correct.
The Serre-Swan theorem gives an equivalence of categories between vector bundles and locally-free $\mathcal{O}_X$-modules of finite rank.
Under this equivalence, the pullback of vector bundles corresponds to the pullback of sheaves of $\mathcal{O}_X$-modules. In particular, for $x: \bullet \hookrightarrow X$ the inclusion of a point, the pullback functor on vector bundles $$x^*: \mathsf{Vect}(X) \to \mathsf{Vect}(\bullet) \cong k(x)\mathsf{Mod}$$ gives you the fibre of a vector bundle, and the pullback functor on sheaves of modules over locally ringed spaces
$$x^*: \mathcal{O}_X\mathsf{Mod} \to  k(x) \mathsf{Mod}$$
given by $\mathcal{F} \mapsto k(x) \otimes_{x^{-1} \mathcal{O}_X} \mathcal{F}$
gives you the fibre of the corresponding sheaf. If $V$ is a vector bundle and $\mathcal{V}$ the corresponding sheaf, then $x^* \mathcal{V}$ agrees with $x^*V$.