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  "body": "\n## Proof\n\nWe prove the two directions separately. Let $\\tau$ be the [[product topology|04S9HX3GAP]] on $X$, and let $\\mathcal{S} = \\set{\\pi_i^{-1}(U) : i \\in I, \\ U \\text{ open in } X_i}$.\n\n1. By definition, $\\tau$ is the topology generated by the [[subbasis|0CHR35YNJQ]] $\\mathcal{S}$. In particular $\\mathcal{S} \\subseteq \\tau$.\n2. Each projection $\\pi_i$ is [[continuous|0SY9ECQ19Q]]. Indeed, for $U$ open in $X_i$, the [[preimage|0FGCFY1AZX]] $\\pi_i^{-1}(U)$ is in $\\mathcal{S} \\subseteq \\tau$.\n3. Let $f$ be continuous. Then each $\\pi_i \\circ f$ is a [[composition of continuous maps|05NK26G0DR]], so it is continuous.\n4. Conversely, let each $\\pi_i \\circ f$ be continuous. Let $\\sigma$ be the set of all $W \\subseteq X$ such that $f^{-1}(W)$ is open in $Z$.\n5. We check that $\\sigma$ is a [[topology|0CNQXVBHCX]] on $X$. First, $f^{-1}(\\varnothing) = \\varnothing$ and $f^{-1}(X) = Z$ are open in $Z$. So $\\varnothing, X \\in \\sigma$.\n6. For a family $(W_k)$ in $\\sigma$, $f^{-1}(\\bigcup_k W_k) = \\bigcup_k f^{-1}(W_k)$, a union of open sets of $Z$. So $\\bigcup_k W_k \\in \\sigma$.\n7. For $W, W' \\in \\sigma$, $f^{-1}(W \\cap W') = f^{-1}(W) \\cap f^{-1}(W')$, which is open in $Z$. So $W \\cap W' \\in \\sigma$.\n8. Let $i \\in I$ and $U$ open in $X_i$. Then $f^{-1}(\\pi_i^{-1}(U)) = (\\pi_i \\circ f)^{-1}(U)$, by the definition of the [[composition|07HF9YCWPP]]. This set is open in $Z$, so $\\pi_i^{-1}(U) \\in \\sigma$.\n9. So $\\sigma$ is a topology on $X$ that contains $\\mathcal{S}$. Since $\\tau$ is the smallest such topology, $\\tau \\subseteq \\sigma$.\n10. Thus $f^{-1}(W)$ is open in $Z$ for each $W \\in \\tau$, and $f$ is continuous.\n\n## Idea\n\nThe sets whose preimages under $f$ are open always form a topology, because preimages respect unions and intersections. If the coordinate maps are continuous, this topology contains the subbasis of the product topology. It then contains the whole product topology, which is the smallest topology containing that subbasis.\n",
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