How do i proof that that the map $ \varphi: Aut(F_n) \to GL_n(\mathbb{Z})$ is homomorpism? I'm trying to proof that a map  $ \varphi: Aut(F_n) \to GL_n(\mathbb{Z})$ is a homomorphism but i can't exactly define which is the function  to show that. 
The map $ \varphi $ is a map with for any $ \alpha \in Aut(F_n)  $ the $(i,j)$-th entry of the matrix $ \varphi(a) $ is the sum of exponents of the letter $x_j$ in the $ a(x_i)$. 
$F_n$ denotes the non-abelian free group generated by $n$ elements

The easiest way to see this is to note that this map factors through an abelianization of $F_n$. Namely, if we quotient by the commutator subgroup then an element $x_{i_1}x_{i_2}\cdots x_{i_n}$ reduces to $x_1^{m_1}x_2^{m_2}\cdots x_n^{m_n}$, where $m_i$ is the sum of the exponents of $x_i$ in the expression. For any automorphism $\alpha:F_n\to F_n$ there is a corresponding automorphism $\alpha':\mathbb{Z}^n\to\mathbb{Z}^n$ where $\alpha'(x_i)$ is the element obtained by combining the powers of each generator. But $\alpha'$ is exactly the element of $\mathrm{GL}_n(\mathbb{Z})$ that $\alpha$ is mapped to.
