 Solution

Identify $T^2$ with the product of two unit circles and use the unit sphere $S^2\subset {\mathbb{R}}^3$. The surjective degree-zero sphere-to-torus map isGiven any torus point, choose its two angle representatives with $x,y\in [-1/2,1/2]$. Then $x^2+y^2\le 1/2$, so $z=\sqrt{1-x^2-y^2}$ completes a preimage on the sphere. Thus the map is surjective. The explicit homotopy $F_t(x,y,z)=(e^{2\pi itx},e^{2\pi ity})$, $0\le t\le 1$, contracts it to a constant map. By homotopy invariance of mapping degree, $\boxed{\mathrm{deg}F=0}$, despite surjectivity.