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**tonio** Let $\displaystyle A=\{a_1,...,a_n\}\,,\,\,B\{b_1,...,b_n\}$ :

1) Suppose f is 1-1 ==> the set $\displaystyle f(A)\subset B$ has as many elements as A (and as B) ==> $\displaystyle f(A)=B$, otherwise B is a finite set having a proper subset with its same cardinality and this is the definition of INFINITE sets ==> f is onto.

2) Suppose now that f is onto ==> $\displaystyle f(A)=B$; if $\displaystyle f(a_i)=f(a_j)\,\,and\,\,a_i\neq a_j$ then $\displaystyle Card(f(A))\leq n-1$, which is impossible since $\displaystyle Card(F(A))=Card (B)=n$, thus it must be that $\displaystyle a_i=a_j$ ==> f is 1-1.

Tonio