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  "body": "\n## Proof\n\nWe check the definition of continuity with preimages of open sets.\n\n1. Let $X$, $Y$ and $Z$ be [[topological spaces|0CNQXVBHCX]], and let $f \\colon X \\to Y$ and $g \\colon Y \\to Z$ be [[continuous|0SY9ECQ19Q]].\n2. Let $W$ be an [[open set|01ZQQDCERP]] of $Z$. We show that the [[preimage|0FGCFY1AZX]] $(g \\circ f)^{-1}(W)$ is open in $X$.\n3. Let $x \\in X$. By the definition of the [[composition|07HF9YCWPP]], $x \\in (g \\circ f)^{-1}(W)$ holds if and only if $g(f(x)) \\in W$.\n4. This holds if and only if $f(x) \\in g^{-1}(W)$, which holds if and only if $x \\in f^{-1}(g^{-1}(W))$.\n5. By steps 3 and 4, $(g \\circ f)^{-1}(W) = f^{-1}(g^{-1}(W))$.\n6. The map $g$ is continuous and $W$ is open in $Z$, so $g^{-1}(W)$ is open in $Y$.\n7. The map $f$ is continuous and $g^{-1}(W)$ is open in $Y$, so $f^{-1}(g^{-1}(W))$ is open in $X$.\n8. By steps 5 and 7, $(g \\circ f)^{-1}(W)$ is open in $X$ for each open set $W$ of $Z$. So $g \\circ f$ is continuous.\n\n## Idea\n\nThe preimage under a composition is the preimage of a preimage, taken in the opposite order.\nEach of the two preimage steps keeps open sets open, so the composition does too.\n",
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  "name": "Proof that the composition of continuous maps is continuous",
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