I am asked to show that every automorphism of the fundamental group of a torus $T=S^1\times{}S^1$ is induced by a homeomorphism $h:T\rightarrow{}T$, which fixes the base point.
What I was thinking is composing rotations in $R^3$ and the antipodal map on $T$, which will give me base-point preserving homeomorphisms of $T$. These homeormorphisms will obviously induce automorphisms of the group $\pi_1(T)$.
But how do I show that all the automorphisms are induced this way? I can see that the order of the group $Aut(\pi_1(T))$ is $8$ and so I have to produce $8$ such homeomorphism! Is there any other elegant way?
Any help is appreciated. Thanks!
Added: As pointed out in the comments, I miscalculated the order of the group $Aut(\pi_1(T))$. But then my original argument is not working at all. Any help?
Think of the torus as $\Bbb{R}^2 / \Bbb{Z}^2$. The map $$ \begin{align} \Bbb{R}^2 &\overset{\normalsize\tilde{\varphi}^{}_A}{\to} \Bbb{R}^2 \\ \begin{bmatrix} s \\ t \end{bmatrix} &\mapsto \begin{bmatrix} as+bt \\ cs+dt \end{bmatrix} \end{align} $$ with $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \in GL_2(\Bbb{Z})$ satisfies $\tilde{\varphi}^{}_A(\Bbb{Z}^2) \subseteq \Bbb{Z}^2$. Therefore, it descends to $$ \begin{align} T &\overset{\normalsize\varphi^{}_A}{\to} T \\ \begin{bmatrix} s \\ t \end{bmatrix} &\mapsto \begin{bmatrix} as+bt \pmod{1} \\ cs+dt \pmod{1} \end{bmatrix} \end{align} $$ For any $A \in GL_2(\Bbb{Z}) \cong \operatorname{Aut}\pi_1(T)$, this is an explicit construction of a self-homeomorphism of $\varphi^{}_A$ of $T$ such that $$ \bigl( \varphi^{}_A \bigr)_* = A. $$