Chapter 6 - Section 6.2 - The Pigeonhole Principle - Exercises - Page 405: 8

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Let f : S $ \rightarrow$ T be a function such that S and T are finite sets with |S| > |T|. Here, number of elements in S are more than the number of elements in T. Since by definition of a function every element should have a unique image, but elements in S are more than elements in T. So, clearly there exists some $s_1$, $s_2$ $\in $ S such that $\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$ f($s_1$)=f($s_2$) i.e., two or more elements of S have same image in T. Thus, f is not one-one.