# Kenneth Rosen Edition 7th Exercise 2.3 Question 72 (Page No. 155)

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Suppose that $f$ is a function from $A$ to $B$, where $A$ and $B$ are finite sets with $|A|=|B|$. Show that $f$ is one-to-one if and only if it is onto.

$\because$ this is a Biconditional proof , we'd have to proof both ways.
1.First, let us prove f is one-to-one $\Rightarrow$ f is onto.
Assume f is not an onto function,$\Rightarrow \exists$ a value in the range which does not have an incoming edge from the domain, but $\because$ |A|=|B|, this cannot happen. $\Rightarrow$ f is onto.

2.1.Now, let us prove f is onto $\Rightarrow$ f is one-to-one .
Assume f is not one-to-one, then |B|<|A|, but we know |A|=|B|, Hence our assumption is false. Therefore f is one-one.

$\therefore$ f is one-to-one $\Leftrightarrow$ f is onto.

Hence proved.

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