Let $T=\mathbb{R}^n/\mathbb{Z}^n$ be the $n$-torus. It is not hard to show that the every automorphism of $T$ as a topological group lifts $T^n$ to an $\mathbb{R}$-linear map $\mathbb{R}^n\to\mathbb{R}^n$ represented by a matrix in $GL_n(\mathbb{Z})$ of determinant 1. (Perhaps we should denote the group of such matrices by $L_n(\mathbb{Z})$, following Serre's joke.) I was trying to show that:
The automorphism group $\operatorname{Aut}_{\mathsf{TopGrp}}(T)$ is a discrete group if we view it as a subspace of $T^T$ with the compact-open topology.
I manage to give a proof, which I indicate below, but I wonder whether there is a proof based on a more elementary observations on matrices. I would appreciate any help. Thanks in advance!
My attempt: Being a covering, $\pi:\mathbb{R}^n\to T$ is a Serre fibration. Since fibrations has the LLP with respect to acyclic cofibrations, it follows that $\pi$ has the LLP with respect to $\operatorname{Aut}_{\mathsf{TopGrp}}(T)\times 0\subset \operatorname{Aut}_{\mathsf{TopGrp}}(T)\times \mathbb{R}^n$. We now consider the following commutative diagram:
$\require{AMScd}$ \begin{CD} \operatorname{Aut}_{\mathsf{TopGrp}}(T)\times 0 @>{\operatorname{pr}_2}>> \mathbb{R}^n\\ @VVV @VVV\\ \operatorname{Aut}_{\mathsf{TopGrp}}(T)\times \mathbb{R}^n@>{(1)}>> T. \end{CD}
The map (1) is given by $(f,x)\mapsto f(\pi(x)).$ Let $\lambda:\operatorname{Aut}_{\mathsf{TopGrp}}(T)\times \mathbb{R}^n\to \mathbb{R}^n$ be a solution to the above lifting problem. Then the transpose of $\lambda $ defines a continuous map $\operatorname{Aut}_{\mathsf{TopGrp}}(T)\to GL_n(\mathbb{R})$. This map assigns to each $f\in \operatorname{Aut}_{\mathsf{TopGrp}}(T)$ the corresponding matrix of $f$, so the claim now follows from the observation that $GL_n(\mathbb{Z})$ is discrete in $GL_n(\mathbb{R}).$ $\blacksquare$