Proving a function is bijection

Asked Jan 30 '18 at 22:32

Active Jan 30 '18 at 23:40

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My problem is: Let $A$ and $B$ finite sets of equal cardinality. Show that if a function $f : A \to B$ is injective then it is also surjective. Similarly show that if $f$ is surjective, then it is also injective.

My thought is since the number of elements in $|A| = |B|$, function $f: A \to B$ is bijection. But my instructor said I have to explain more. Please help me with this. I don't know how a injective function leads to a surjective function and vice versa. Thank you.

edited Jan 30 '18 at 23:23

Mustafa

1,590

asked Jan 30 '18 at 22:32

JTJung

The nice thing about finite cardinality is you can do standard inductions with them. – Noah Riggenbach Jan 30 '18 at 22:35
2

when you wrote "if |A| = |B|, then f: A->B is bijection" is just to rewrite the same question, is not its proof – yemino Jan 30 '18 at 22:35
See https://math.stackexchange.com/a/2189220/589. – lhf Jan 30 '18 at 22:36
@vnsoshi, Just because they both have finite cardinality, does not mean that they have the same cardinality. Injectivity would give you that $|A| \leq |B|.$ – Sahiba Arora Jan 30 '18 at 22:36

Proving a function is bijection

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