Toeplitz operator: $T_f: L^2(S^1) \rightarrow L^2(S^1)$ is given by composition $PM_f$, where $M_f:u \mapsto fu$ is the multiplication map by continuous function $f:S^1 \rightarrow \Bbb C$ and $P:L^2(S^1) \rightarrow H(S^1)$ is the projection map on the Hardy space.
It is claimed that
$PM_fP + (1-P)$ has the same index as $T_f$.
Index of operator $A$ is defined as $ind(A):= \dim \ker A - \dim \text{coker } A$ supposing it exists.
How is this statement true?
I know that $ind(F \circ G) = ind(F) + ind(G)$ (supposing both exists), so precompose the map by $id=P+(1-P)$ and we should have the formula $PM_fP + PM_f(1-P)$...
Reference page 12 line 5.
You have to look at the operator $P M_f P + (1-P)$ directly. Since $P$ is the orthogonal projection onto Hardy space $H^2(S^1)$, it follows that $1-P$ is the orthogonal projection onto its orthogonal complement $H^2(S^1)^\perp$. Thus, if you think of $L^2(S^1)$ as the orthogonal direct sum $L^2(S^1) = H^2(S^1) \oplus H^2(S^1)^\perp$, then you can view $P M_f P + (1-P)$ as the block-diagonal operator $$ PM_f P + (1-P) = \begin{pmatrix} T_f \vert_{H^2(S^1)} & 0 \\ 0 & 1_{H^2(S^1)^\perp} \end{pmatrix}, $$ which makes it clear that $PM_f P + (1-P) : L^2(S^1) \to L^2(S^1)$ has the same kernel and the same cokernel as the Toeplitz operator $T_f : H^2(S^1) \to H^2(S^1)$, and hence the same index. Really, all you're doing is extending $T_f$ from $H^2(S^1)$ to $L^2(S^1)$ as simply as possible while preserving its index theory.