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  "body": "\n## Statement\n\nLet $X$ be a [[topological space|0CNQXVBHCX]], let $\\sim$ be an equivalence relation on $X$, and give $X / {\\sim}$ the [[quotient topology|0V8046R0P9]], with quotient map $q \\colon X \\to X / {\\sim}$.\nLet $Y$ be a topological space and let $g \\colon X / {\\sim} \\to Y$ be a function.\nThen $g$ is [[continuous|0SY9ECQ19Q]] if and only if the [[composition|07HF9YCWPP]] $g \\circ q \\colon X \\to Y$ is continuous.\n\n## Remarks\n\nProof of the converse: for $V$ open in $Y$, the set $q^{-1}(g^{-1}(V)) = (g \\circ q)^{-1}(V)$ is open in $X$. Thus $g^{-1}(V)$ is open in the quotient.\nSo each continuous map $f \\colon X \\to Y$ that is constant on each class gives a unique continuous map $g$ with $f = g \\circ q$.\nFor example, $t \\mapsto (\\cos 2\\pi t, \\sin 2\\pi t)$ is constant on the classes of $\\mathbb{R} / \\mathbb{Z}$, so it gives a continuous map from $\\mathbb{R} / \\mathbb{Z}$ to the circle.\n\n## Explanations\n\n### L1 human\n\nTo define a continuous map on a quotient space, define a continuous map on the original space that takes the same value on glued points.\nThe map then passes to the quotient.\n",
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