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  "body": "\n## Statement\n\nLet $X$ and $Y$ be [[Banach spaces|06GZ956657]] over $\\R$ or $\\C$.\nLet $T \\colon X \\to Y$ be a [[continuous linear map|03TM135G0N]] that is [[bijective|0GG9TVW90M]].\nThen the [[inverse|0J2WAR83WP]] $T^{-1} \\colon Y \\to X$ is continuous, that is, there is a constant $C \\ge 0$ with $\\norm{T^{-1} y} \\le C \\norm{y}$ for all $y \\in Y$.\n\n## Remarks\n\nThe theorem follows from the open mapping theorem: a surjective continuous linear map between Banach spaces sends open sets to open sets.\nThat theorem is proved with the [[Baire category theorem|0A7KGAYHGV]] applied to the cover $Y = \\bigcup_k \\overline{T(B(0, k))}$.\nFor a bijection, openness of $T$ is exactly continuity of $T^{-1}$.\n\nA common application: if a vector space is complete for two norms $\\norm{\\cdot}_1$ and $\\norm{\\cdot}_2$ with $\\norm{x}_2 \\le c \\norm{x}_1$, then the two norms are equivalent.\nIndeed, the identity map is a continuous bijection from the first norm to the second.\nCompleteness of $Y$ is necessary.\nOn $\\ell^1$ the identity map from $(\\ell^1, \\norm{\\cdot}_1)$ to $(\\ell^1, \\norm{\\cdot}_\\infty)$ is a continuous linear bijection, because $\\norm{x}_\\infty \\le \\norm{x}_1$.\nIts inverse is not bounded: $x = e_1 + \\cdots + e_n$ has $\\norm{x}_\\infty = 1$ and $\\norm{x}_1 = n$.\nHere $(\\ell^1, \\norm{\\cdot}_\\infty)$ is not complete.\n\n## Explanations\n\n### L1 human\n\nIf a continuous linear map between complete normed spaces is one-to-one and onto, then undoing it is also continuous.\nSo such a map is an isomorphism of Banach spaces, with no extra checks.\nSmall changes in the output come from small changes in the input.\n\n### L2 human\n\nThe theorem is used to prove stability of solutions: if an equation $Tx = y$ has exactly one solution for each $y$, the solution depends continuously on $y$.\nIt is equivalent, by short arguments, to the open mapping theorem and to the [[closed graph theorem|GXM8BX8YM7]].\nTogether with the [[uniform boundedness principle|YG0042V8CJ]] these results all rest on the [[Baire category theorem|0A7KGAYHGV]].\n",
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