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  "body": "\n## Definition\n\nLet $\\mathcal{C}$ and $\\mathcal{D}$ be categories.\nA [[functor|28NQ0KRHD3]] $F \\colon \\mathcal{C} \\to \\mathcal{D}$ is an *equivalence of categories* if there are a functor $G \\colon \\mathcal{D} \\to \\mathcal{C}$ and [[natural isomorphisms|KYCPCC57FT]] $\\eta \\colon 1_{\\mathcal{C}} \\Rightarrow G \\circ F$ and $\\varepsilon \\colon F \\circ G \\Rightarrow 1_{\\mathcal{D}}$.\nThen $G$ is a *quasi-inverse* of $F$, and $\\mathcal{C}$ and $\\mathcal{D}$ are *equivalent*, written $\\mathcal{C} \\simeq \\mathcal{D}$.\n\n## Remarks\n\nIf moreover $G \\circ F = 1_{\\mathcal{C}}$ and $F \\circ G = 1_{\\mathcal{D}}$, then $F$ is an *isomorphism of categories*, a much stronger condition.\nExample: let $K$ be a field and let $\\mathbf{Mat}_K$ have the natural numbers as objects, the $n \\times m$ matrices over $K$ as morphisms $m \\to n$, and matrix multiplication as composition.\nThe functor $\\mathbf{Mat}_K \\to$ (finite-dimensional [[vector spaces|000AYPA9AJ]] over $K$) sending $n$ to $K^n$ and a matrix to its linear map is an equivalence.\nA quasi-inverse chooses a basis of each space $V$ and sends $V$ to $\\dim V$ and a linear map to its matrix in the chosen bases.\nNon-example: let $\\mathbf{1}$ have one object and one morphism, and let $\\mathbf{2}$ have two objects and only identity morphisms.\nNo functor $F \\colon \\mathbf{1} \\to \\mathbf{2}$ is an equivalence.\nBoth objects $X$ of $\\mathbf{2}$ would need an isomorphism $F(G(X)) \\cong X$.\nBut $F(G(X))$ is the single object in the image of $F$, and the two objects of $\\mathbf{2}$ are not isomorphic.\nEquivalence of categories is reflexive, symmetric and transitive.\n",
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