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  "body": "\n## Proof\n\nWe check the definition of surjectivity directly.\n\n1. Let $f \\colon A \\to B$ and $g \\colon B \\to C$ be [[surjective|0XKH881BZ7]] functions.\n2. Let $c \\in C$. Since $g$ is surjective, there is $b \\in B$ with $g(b) = c$.\n3. Since $f$ is surjective, there is $a \\in A$ with $f(a) = b$.\n4. By the definition of [[composition|07HF9YCWPP]] and steps 2 and 3, $(g \\circ f)(a) = g(f(a)) = g(b) = c$.\n5. So each $c \\in C$ equals $(g \\circ f)(a)$ for some $a \\in A$. Hence $g \\circ f$ is surjective.\n\n## Idea\n\nTo hit a target $c$, first find a point $b$ that $g$ sends to $c$. Then find a point $a$ that $f$ sends to $b$. That $a$ is sent to $c$ by $g \\circ f$.\n",
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