= 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> is
$$
\boxed{F(x,y,z)=(e^{2\pi ix},e^{2\pi iy}).}
$$
Given any torus point, choose its two angle representatives with $x,y\in[-1/2,1/2]$. Then $x^2+y^2\leq1/2$, so $z=\sqrt{1-x^2-y^2}$ completes a preimage on the sphere. Thus the map is surjective. The explicit <homotopy> $F_t(x,y,z)=(e^{2\pi itx},e^{2\pi ity})$, $0\leq t\leq1$, contracts it to a constant map. By <homotopy invariance of mapping degree>, $\boxed{\deg F=0}$, despite surjectivity.
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