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  "body": "\n## Definition\n\nLet $X$ and $Y$ be [[topological spaces|0CQHFPGJ45]] and let $x \\in X$.\nConsider pairs $(U, f)$, where $U$ is an open [[neighbourhood|0ZCY0Z80B6]] of $x$ and $f \\colon U \\to Y$ is a map.\nTwo such pairs $(U, f)$ and $(V, g)$ are *equivalent at $x$* if there is an open neighbourhood $W \\subseteq U \\cap V$ of $x$ with $f(w) = g(w)$ for all $w \\in W$.\nThis is an [[equivalence relation|0G8VEZ3WTK]].\nThe *germ* of $(U, f)$ at $x$ is its [[equivalence class|0W95W33CMP]].\n\n## Remarks\n\nOne often restricts to continuous, smooth or holomorphic maps, and writes $[f]_x$ for the germ.\nThe value $f(x)$ depends only on the germ, and so do derivatives of all orders at $x$ when $f$ is smooth.\nMore generally, let $\\mathcal{F}$ be a filter on a set $S$.\nTwo maps on $S$ have the same germ along $\\mathcal{F}$ if they agree on some member of $\\mathcal{F}$.\nThe case above uses the neighbourhood filter of $x$.\n\nExample: on $\\R$, the maps $f(t) = t$ and $g(t) = \\abs{t}$ have the same germ at $1$, since they agree on $W = (0, 2)$.\n\nNon-example: on $\\R$, let $f = 0$ and let $g(t) = e^{-1/t^2}$ for $t \\ne 0$, with $g(0) = 0$.\nBoth are smooth, and all their derivatives agree at $0$.\nYet $g(t) \\ne 0$ for every $t \\ne 0$, so they agree on no neighbourhood of $0$, and their germs at $0$ differ.\n",
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