TITLE: Automorphisms of a torus
QUESTION [5 upvotes]: Why is the set of automorphism of a n-dimensional torus T (also denoted $T^n$) is $GL(n,\mathbb{Z})$ i.e the set of invertible matrices with integral coefficients?
In the book by Brocker and Dieck- Representations of Compact Lie Groups, the crucial step for proving this goes as follows-

If there is a homomorphism $\phi:T^n\to S^1$, then it induces the diagram
$$\require{AMScd} \begin{CD}  \mathbb{Z}^n @>{\hookrightarrow}>> \mathbb{R}^n @>{D\phi}>> \mathbb{R}\\@VVV @VVV @VVV \\  \mathbb{Z} @>{\hookrightarrow}>> T^n @>{\phi}>> S^1 \end{CD}$$
which gives the result that $$D\phi(v_1,...,v_n)= \sum n_iv_i$$ where $n_i \in \mathbb{Z}$; where the vertical maps are projections.

Q.1. How is the above formula for $D\phi$ inferred?
Other than that argument, I have managed to show that $GL(n,\mathbb{Z})\subset Aut(T^n)$. But how to show the other inclusion? That is,
Q.2. Given an automorphism $\phi$ of $T^n= \mathbb{R}^n/\mathbb{Z}^n$, show that there exist a map $\Phi$ $\in$ $GL(n,\mathbb{Z})$ such that $\pi \circ\Phi = \phi$ , where $\pi:R^n\to T^n$ is the projection.
Thanks in advance!

REPLY [2 votes]: To put it in a concise answer:
A.1: If for some $i$, $n_i \in \mathbb{R}$ but not in $\mathbb{Z}$, then the corresponding unit vector will map to $$D\phi(0,...,0,e_i,0,..0)=\sum_{i\neq j} n_j. 0 + n_i.e_i = n_i.e_i$$ which will not be sent to $\mathbb{Z}$. Hence the entries of the matrix must be integers.
A.2: (As detailed above) Any homomorphism $\phi \in Aut(T^n)$ can be lifted to $Aut(\mathbb{R}^n)$, because of the lifting property of universal cover. (Since $\mathbb{R}^n$ is the universal cover of $T^n$). Therefore $$Aut(T^n)\subset Aut(\mathbb{R}^n)$$
Hence the automorphism is given by a matrix.