= Solution
The <Fredholm index> measures the finite-dimensional obstruction to inverting an operator. For a <bounded operator> $T:H_1\to H_2$ between <Hilbert spaces>, the <Fredholm> conditions are that its <range of a bounded linear operator> is closed and its <kernel> and <cokernel> are finite-dimensional. The <orthogonal complement> of the range is $\ker T^*$, so
$$
\boxed{\operatorname{ind}T=\dim\ker T-\dim\ker T^*.}
$$
An <invertible linear map> has <Fredholm index> zero; the converse needs an additional finite-dimensional correction, since an operator can have equally large <kernel> and <cokernel>.
First construct an inverse up to <finite-rank operators>. On $(\ker T)^\perp$, the restriction of $T$ is a bounded bijection onto its closed range, with bounded inverse by the <bounded inverse theorem>. Extend that inverse by zero on $(\operatorname{ran}T)^\perp$, obtaining $S:H_2\to H_1$. Then
$$
ST=I-P_{\ker T},\qquad TS=I-P_{\ker T^*}.
$$
Conversely, suppose $ST-I$ and $TS-I$ are <compact operators>. On $\ker T$, the first identity expresses the identity as a <compact operator>, so this <kernel> is finite-dimensional. If $T$ were not bounded below on $(\ker T)^\perp$, there would be unit <vectors> $x_j$ in that complement with $Tx_j\to0$. Write $ST=I+K$. <Compactness> of $K$ gives a subsequence for which $Kx_j$ converges, hence $x_j=STx_j-Kx_j$ converges to a unit <vector> $x$ in $(\ker T)^\perp$ with $Tx=0$, a contradiction. This lower bound makes the range closed. Applying the analogous argument to $T^*$, using $S^*T^*=I+K'^*$, proves that its <kernel> is finite-dimensional. This proves the <Atkinson theorem>: a <bounded operator> is <Fredholm> exactly when it has an inverse modulo <compact operators>. For operators on one <Hilbert space> this says that its image in the <Calkin algebra> is invertible.
The <Fredholm index> is stable. Decompose the domain as $N\oplus N^\perp$, where $N=\ker T$, and the target as $C\oplus R$, where $C=\ker T^*$ and $R=\operatorname{ran}T$. A sufficiently small <norm> perturbation has block <matrix>
$$
T+E=\begin{pmatrix}a&b\\c&D\end{pmatrix},\qquad D:N^\perp\longrightarrow R\text{ invertible}.
$$
Multiplying by invertible triangular block <matrices> reduces this <matrix> to the <direct sum> of $D$ and the <Schur complement> $a-bD^{-1}c:N\to C$. The latter's <kernel> <dimension> minus <cokernel> <dimension> is $\dim N-\dim C$, whatever its <rank>. Therefore the Fredholm set is open and the index is locally constant. If $K$ is compact, the same parametrix works modulo <compact operators> for every $T+tK$, $0\leq t\leq1$. Local constancy along this path gives
$$
\operatorname{ind}(T+K)=\operatorname{ind}T.
$$
For <Fredholm operators> $A:H_1\to H_2$ and $B:H_2\to H_3$, multiplication of parametrices shows that $BA$ is Fredholm. Its index is additive. Here is a direct proof, rather than an assertion of additivity. There is an <exact sequence> of finite-dimensional spaces
$$
0\longrightarrow\ker A\longrightarrow\ker BA
\xrightarrow{A}\ker B\longrightarrow\operatorname{coker}A
\xrightarrow{B}\operatorname{coker}BA
\longrightarrow\operatorname{coker}B\longrightarrow0.
$$
The map from $\ker B$ takes a <vector> to its class modulo $\operatorname{ran}A$; the next takes the class of $y$ to that of $By$; the last is the quotient map. For example, a class $y+\operatorname{ran}A$ lies in the <kernel> of the middle map exactly when $By=BAx$ for some $x$, so $y-Ax\in\ker B$. This verifies the only less immediate exactness assertion; the others follow directly from their definitions. Alternating the <dimensions> gives
$$
\operatorname{ind}(BA)=\operatorname{ind}B+\operatorname{ind}A.
$$
Also $\operatorname{ind}T^*=-\operatorname{ind}T$, and indices add under <direct sums>. For $I+K$ with $K$ compact, the index is zero, so <injectivity> is equivalent to <surjectivity>. More generally $Tx=f$ is solvable exactly when $f$ is orthogonal to $\ker T^*$: these are the finitely many compatibility conditions in the <Fredholm alternative>.
<Toeplitz operators> exhibit how this analytic index detects a topological winding. Let $H^2\subset L^2(S^1)$ be the <Hardy space of the circle>, spanned by $1,z,z^2,\ldots$, and let $P$ be its <orthogonal projection>. For a continuous symbol $g$, define $T_g=P M_g|_{H^2}$, where $M_g$ is multiplication by $g$. Then $\|T_g\|\leq\|g\|_\infty$. The product defect is
$$
T_fT_g-T_{fg}=-P M_f(I-P)M_g|_{H^2}.
$$
For <trigonometric polynomials> the crossing of negative and nonnegative <Fourier modes> has finite <rank>. Uniform approximation of continuous symbols by <trigonometric polynomials>, together with the displayed <norm> estimate, therefore makes this defect a <compact operator> for all continuous $f,g$. If $g$ is nowhere zero, $1/g$ is continuous and $T_{1/g}$ is a two-sided parametrix modulo <compact operators>. Thus $T_g$ is Fredholm.
The converse is instructive. If $g(\zeta)=0$ at $\zeta\in S^1$, consider the normalized Hardy <kernels>
$$
k_{r,\zeta}(z)=\frac{\sqrt{1-r^2}}{1-r\overline\zeta z},\qquad 0<r<1.
$$
Their <norms> are one, their individual <Fourier coefficients> tend to zero as $r\uparrow1$, and hence they converge weakly to zero. Their squared moduli are the <Poisson kernel>, so continuity of $g$ gives
$$
\|T_g k_{r,\zeta}\|_2^2\leq\|gk_{r,\zeta}\|_2^2
=\int_{S^1}|g(z)|^2|k_{r,\zeta}(z)|^2\,\frac{|dz|}{2\pi}\longrightarrow0.
$$
A <Fredholm operator> is bounded below off its finite-dimensional <kernel>. The projection of these weakly vanishing <vectors> onto that <kernel> tends to zero, contradicting this lower bound. Thus \b[a continuous-symbol <Toeplitz operator> is Fredholm exactly when its symbol never vanishes]. Applying the same argument to $g-\lambda$ gives its <essential spectrum> as $g(S^1)$, with essential <spectrum> defined by failure of the Fredholm property.
For nonvanishing $g$, let $m=\operatorname{wind}(g,0)$. Lifting a continuous argument of $g(e^{it})$ on $0\leq t\leq2\pi$ gives an argument change $2\pi m$. After division by $z^m$, its argument has matching endpoints, so there is a <continuous function> $h:S^1\to\mathbb C$ with $g(z)=z^m e^{h(z)}$. The nonvanishing-symbol homotopy $g_t=z^m e^{th}$ reduces the index to that of $T_{z^m}$. For $m\geq0$ this is the $m$th power of the <unilateral shift operator>, whose <kernel> is zero and whose range has codimension $m$. For $m<0$ it is the $(-m)$th power of the adjoint shift, whose <kernel> has <dimension> $-m$ and whose range is all of $H^2$. Consequently
$$
\boxed{\operatorname{ind}T_g=-\operatorname{wind}(g,0).}
$$
The minus sign comes from counting missing nonnegative <Fourier modes>. This is a concrete bridge between the <Fredholm index>, stability under compact perturbations, and the <winding number> of a loop.
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