= Solution
On the <Hardy space of the circle> $H^2(\mathbb T)\subset L^2(\mathbb T)$, let $P_+$ be the projection onto nonnegative Fourier modes and $T_f=P_+M_f|_{H^2}$. The <Toeplitz index theorem for continuous symbols> states: for continuous $f$, $T_f$ is <Fredholm> exactly when $f$ never vanishes, and then
$$
\boxed{\operatorname{ind}T_f=-\operatorname{wind}(f,0),}
$$
with counterclockwise winding positive.
First, $T_fT_g-T_{fg}=-P_+M_f(I-P_+)M_gP_+$ is compact. For Laurent-polynomial symbols the negative-mode crossing involves only finitely many modes and is finite rank. Uniform approximation and $\|T_f\|\leq\|f\|_\infty$ extend this to continuous symbols. If $f$ is nonvanishing, $T_{1/f}$ is therefore a two-sided parametrix, and part (b) gives the <Fredholm> property.
Writing $n=\operatorname{wind}(f,0)$, a continuous argument along the parametrized circle increases by $2\pi n$. Removing this increase gives a periodic continuous logarithm $h$ of $z^{-n}f(z)$, so $f=z^ne^h$. The symbols $z^ne^{th}$ give a continuous <Fredholm> path from $z^n$ to $f$. Part (c) makes their indices equal. For $n\geq0$, $T_{z^n}$ is the $n$th unilateral shift, with zero kernel and $n$-dimensional cokernel. For negative $n$ it is the corresponding backward shift, with kernel dimension $-n$ and zero cokernel. In both cases the index is $-n$.
For necessity, if $f(\zeta)=0$ use the normalized Hardy kernels
$$
k_{r,\zeta}(z)=\frac{\sqrt{1-r^2}}{1-r\overline\zeta z},\qquad r\uparrow1.
$$
They have norm one and converge weakly to zero, as their inner products with each Fourier basis vector tend to zero. Their squared moduli are Poisson kernels concentrating at $\zeta$, so continuity gives $\|fk_{r,\zeta}\|_2\to0$ and hence $\|T_fk_{r,\zeta}\|_2\to0$. A <Fredholm> parametrix would imply $k_{r,\zeta}=BT_fk_{r,\zeta}+Kk_{r,\zeta}\to0$ in norm, because a <compact operator> takes this weakly null bounded family to zero in norm. This contradicts unit norm, proving the converse.
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