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  "body": "\n## Proof\n\nWe check the definition of injectivity directly.\n\n1. Let $f \\colon A \\to B$ and $g \\colon B \\to C$ be [[injective|0PVG1CRAJN]] functions.\n2. Let $a, a' \\in A$ with $(g \\circ f)(a) = (g \\circ f)(a')$.\n3. By the definition of [[composition|07HF9YCWPP]], step 2 says $g(f(a)) = g(f(a'))$.\n4. Since $g$ is injective and $f(a), f(a') \\in B$, step 3 gives $f(a) = f(a')$.\n5. Since $f$ is injective, step 4 gives $a = a'$.\n6. So $(g \\circ f)(a) = (g \\circ f)(a')$ implies $a = a'$ for all $a, a' \\in A$. Hence $g \\circ f$ is injective.\n\n## Idea\n\nTo go back from an equal output of $g \\circ f$ to equal inputs, peel off one function at a time. Injectivity of $g$ removes $g$, and then injectivity of $f$ removes $f$.\n",
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