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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.

I want to know my proof is right or wrong?

Proof: by contradiction

f[A] is image of function f.

Let f is one to one but not onto.

f is one to one =  |f[A]|  = |A| and

because its not onto (these exist some x other than f[A])

∃x (x∉ f[A] and x∈B) $\rightarrow$ |B|>|f[A]|  

so, |B|> |A| (which contradicts |A|=|B|)

So f is onto.

Let f is onto but not one to one

f is onto  =  |f[A]|  = |B| and

not one to one  : |B| < |A| (which contradicts |A|=|B|)

so f is  one to one
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