Monodromy of local system is well defined I want to understand, why local systems (= locally constant sheafs) with values in a vector space $V$ define a monodromy representation $\rho: \pi_1(X) \to \operatorname {GL}(V)$.
I already know, that any local system on the interval $I=[0,1]$ and on $I^2$ is constant.
So let $\mathcal F$ be a local system on $X$ and $\gamma: I \to X$ a loop with base point $x_0$. Then $\gamma^{-1} \mathcal F$ is a local system on $I$ and thus constant. Furthermore, the stalks of a inverse image sheaf are given as 
$$(\gamma^{-1} \mathcal F)_t = \mathcal F_{\gamma(t)}.$$
As for a constant sheaf the restriction map from global sections to stalks is an isomorphism, we get an isomorphism 
$$\mathcal F_{x_0} = (\gamma^{-1} \mathcal F)_0 \simeq \Gamma(\gamma^{-1} \mathcal F, I) \simeq (\gamma^{-1} \mathcal F)_1 =  \mathcal F_{x_0}. $$
Why does this isomorphism only depend on the homotpy class of $\gamma$?
Let $H: I \times I \to X$ be a homotopy between $\gamma$ and $\gamma'$, both loops with base point $x_0$.
Then again, $H^{-1} \mathcal F$ is a constant sheaf, and we obtain isomorphisms
$$\mathcal F_{x_0} = (H^{-1} \mathcal F)_{(0,0)} \simeq \Gamma(H^{-1} \mathcal                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                    F, I^2) \simeq (H^{-1} \mathcal F)_{(1,0)}= \mathcal F_{x_0}$$
and 
$$\mathcal F_{x_0} = (H^{-1} \mathcal F)_{(0,0)} \simeq \Gamma(H^{-1} \mathcal F, I^2) \simeq (H^{-1} \mathcal F)_{(0,1)}= \mathcal F_{x_0}.$$
I need to show that one of the isomorphisms above is the identity and the other is the one given by $\gamma$.
 A: Saying $H^{-1}\mathcal{F}$ is a constant sheaf likely assumes what you want to prove: the homotopy-invariance of monodromy. For how else do you show that a sheaf on $[0,1]^2$ is trivial? 
Suppose $\gamma$ and $\gamma'$ are loops based at $x$, homotopic through $H$ (i.e. $H(t,0) = \gamma(t)$ and $H(t,1) = \gamma'(t)$). Then $I \times I$ has a cover by open sets $H^{-1}(U)$ where $\mathcal{F}$ is constant on $U$. Picking a Lebesgue number for this covering, we may subdivide $I \times I$ into smaller rectangles $[a_{k-1},a_k] \times [b_{\ell-1},b_\ell]$ covered by open sets of the above form, where $0 = a_0 < a_1 < \cdots < a_n = 1$ and $0 = b_0 < b_1 < \cdots < b_m = 1$. Let $\gamma_\ell$ be the path given by $\gamma_\ell(t) = H(t, b_\ell)$, so that $\gamma_0 = \gamma$ and $\gamma_m = \gamma'$. We will show that the monodromy action of $\gamma_\ell$ equals the monodromy action of $\gamma_{\ell+1}$ for all $\ell$.
Given paths $\gamma_\ell$ and $\gamma_{\ell+1}$, consider the path $\gamma_\ell^k$ which first traverses $H(t,b_\ell)$ for $t \leq a_k$, then traverses $H(a_k,s)$ for $s \in [b_\ell, b_{\ell+1}]$, then traverses $H(t, b_{\ell+1})$ for $t \geq a_k$. Then $\gamma^k_\ell$ and $\gamma^{k+1}_\ell$ only differ by traversing the boundary of the square $[a_k,a_{k+1}]\times[b_k,b_{k+1}]$ on either the left and bottom or the right and top sides. But $\mathcal{F}$ is constant on that rectangle, so $\gamma_\ell^k$ and $\gamma_\ell^{k+1}$ induce the same monodromy map. Hence $\gamma_\ell = \gamma\ell^0$ and $\gamma_{\ell+1} = \gamma_\ell^{n}$ induce the same monodromy map, as desired.

Edit: Let me give a second proof, using that every element of a stalk of a local system on $[0,1]$ or $[0,1]^2$ extends to a unique section. Let $H : [0,1]^2 \to X$ be our homotopy. We have $H(0,s) = H(1,s) = x_0$, $H(-,0) = \gamma$, and $H(-,1) = \gamma'$. Given $f \in H^{-1}\mathcal{F}_{(0,0)}$, we may extend $f$ to a unique section $\tilde f$ of $H^{-1}\mathcal{F}$. Restricting $\tilde f$ to the sides of the square gives sections of $\mathcal{F}$ over the sides of the square. Since the sections of a local system on an interval are unique, we conclude that on the sides of the rectangle mapping to $x_0$, $\tilde f_{(t,0)} = \tilde f_{(t',0)} = f$ and $\tilde f_{(t,1)} = \tilde f_{(t',1)}$ for all $t,t' \in [0,1]$ (since there is a section given by pulling back $x_0 \mapsto f$). By uniqueness again, $\tilde f_{(0,1)} = \rho(\gamma)(f)$, and $\tilde f_{(1,1)} = \rho(\gamma')(\tilde f_{(1,0)}) = \rho(\gamma')(f)$. Thus, $\rho(\gamma)(f) = \rho(\gamma')(f)$. 
Another way of describing this proof is that the homotopy enforces a commutative diagram
$$ \require{AMScd} \begin{CD} 
 H^{-1}\mathcal{F}_{(0,0)} @>\rho(\gamma)>> H^{-1}\mathcal{F}_{(0,1)} \\
 @V{id}VV @V{id}VV \\
 H^{-1}\mathcal{F}_{(1,0)} @>\rho(\gamma')>> H^{-1}\mathcal{F}_{(1,1)}
\end{CD}
$$
by factoring all maps concerned through $H^{-1}\mathcal{F}_{(0,0)} \to \Gamma( H^{-1}\mathcal{F})$, the map of extending to a section, followed by restriction of a section.
