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  "body": "\n## Proof\n\nWe prove the two directions separately. Let $\\tau$ be the topology of $X$.\n\n1. The quotient map $q$ is [[continuous|0SY9ECQ19Q]]. Indeed, if $V$ is open in $X / {\\sim}$, then $q^{-1}(V) \\in \\tau$ by the definition of the [[quotient topology|0V8046R0P9]].\n2. Let $g$ be continuous. Then $g \\circ q$ is a [[composition of continuous maps|05NK26G0DR]], so it is continuous.\n3. Conversely, let $g \\circ q$ be continuous, and let $V$ be open in $Y$.\n4. By the definition of the [[composition|07HF9YCWPP]], $x \\in (g \\circ q)^{-1}(V)$ if and only if $g(q(x)) \\in V$. This holds if and only if $q(x) \\in g^{-1}(V)$, that is, $x \\in q^{-1}(g^{-1}(V))$.\n5. So the [[preimages|0FGCFY1AZX]] agree: $q^{-1}(g^{-1}(V)) = (g \\circ q)^{-1}(V)$. This set is in $\\tau$ because $g \\circ q$ is continuous.\n6. By the definition of the quotient topology, $g^{-1}(V)$ is open in $X / {\\sim}$.\n7. So the preimage under $g$ of each open set of $Y$ is open, and $g$ is continuous.\n\n## Idea\n\nA set in the quotient is open exactly when its preimage under $q$ is open. The preimage under $q$ of $g^{-1}(V)$ is the preimage of $V$ under $g \\circ q$. So continuity of $g \\circ q$ is exactly what is needed for $g$.\n",
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