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  "body": "\n## Proof\n\nWe argue by contradiction.\n\n1. Let $A$ and $B$ be finite sets, let $f : A \\to B$ be a function, and let $n$ be a natural number with $|B| \\cdot n \u003c |A|$.\n2. For $b \\in B$, write $A_b = f^{-1}(b) = \\set{a \\in A : f(a) = b}$. Each $A_b$ is a subset of the finite set $A$, so it is finite.\n3. Suppose, for a contradiction, that $|A_b| \\le n$ for every $b \\in B$.\n4. The sets $A_b$ for $b \\in B$ are pairwise disjoint. Indeed, if $a \\in A_b \\cap A_{b'}$, then $b = f(a) = b'$.\n5. Their union is $A$, because each $a \\in A$ lies in $A_{f(a)}$.\n6. By the [[rule of sum|0EE2V3CHFK]], applied to the $|B|$ pairwise disjoint finite sets $A_b$, we get $|A| = \\sum_{b \\in B} |A_b|$.\n7. By step 3, each term of this sum is at most $n$, and there are $|B|$ terms. So $|A| \\le |B| \\cdot n$.\n8. This contradicts the hypothesis $|B| \\cdot n \u003c |A|$ of step 1.\n9. Hence some $b \\in B$ has $|f^{-1}(b)| \u003e n$, that is, its fibre has [[cardinality|05099E1NMJ]] greater than $n$.\n\n## Idea\n\nThe fibres of $f$ split $A$ into $|B|$ disjoint pieces, so their sizes add up to $|A|$. If every piece had at most $n$ elements, the total would be at most $|B| \\cdot n$, which is too small.\n",
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