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  "body": "\n## Statement\n\nLet $X$ and $Y$ be [[Banach spaces|06GZ956657]], and let $T \\colon X \\to Y$ be a [[bounded linear map|03TM135G0N]].\nThen $T$ is a [[Fredholm operator|8HWGKWD6ZM]] if and only if there is a bounded linear map $S \\colon Y \\to X$ such that $ST - I_X$ and $TS - I_Y$ are [[compact operators|M2NEP1K5AE]].\n\n## Remarks\n\nIf $T$ is Fredholm, choose closed complements $X = \\ker T \\oplus X_0$ and $Y = T(X) \\oplus Y_0$, with $Y_0$ finite-dimensional.\nThen $T$ maps $X_0$ bijectively onto $T(X)$, and $S$ can be taken as its inverse on $T(X)$ and $0$ on $Y_0$.\nWith this choice, $I_X - ST$ and $I_Y - TS$ are projections onto $\\ker T$ and $Y_0$, so they are even [[finite-rank operators|Z8FABSYQTS]].\nConversely, if $ST = I_X + K$ with $K$ compact, then $\\ker T \\subseteq \\ker(I_X + K)$ is finite-dimensional, and $TS = I_Y + K'$ shows that $T(X) \\supseteq (I_Y + K')(Y)$ has finite codimension.\nFor a Hilbert space $H$, the theorem says that $T \\in B(H)$ is Fredholm if and only if its image is invertible in the [[Calkin algebra|6FE7JZ2VF1]] $B(H)/K(H)$.\nExample: the right shift $S$ on $\\ell^2$ and the left shift $S^*$ satisfy $S^* S = I$ and $S S^* = I - P$, where $P$ is the rank-one projection onto the first coordinate, so $S$ is Fredholm.\n\n## Explanations\n\n### L1 human\n\nA Fredholm operator is one that can be inverted up to a compact error.\nThe theorem says that this is exactly the condition of finite-dimensional kernel and cokernel.\nSo Fredholm operators are the operators that are invertible once one ignores compact operators.\n\n### L2 human\n\nThe theorem turns Fredholm theory into algebra: Fredholm operators are the preimages of the invertible elements of a quotient algebra.\nFrom this view, it follows at once that compact perturbations of Fredholm operators are Fredholm, and that products of Fredholm operators are Fredholm.\nIt also leads to the essential spectrum of an operator $T$ on a Hilbert space, the spectrum of its image in the Calkin algebra.\n",
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