Prove, if is surjective function, then is injective function.

Asked Oct 14 '18 at 00:16

Active Oct 14 '18 at 00:30

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The question would be:

Let – finite set, and : → . Prove, if is a surjective function, then is an injective function.

Proposed solution:

Consider f is not an injective function. Then, it exist a, b: a not equal to b, f(a) = f(b) This means, that at least two different elements become the same after applying f. As U is finite, at least one element in U have an empty preimage. This is an contradiction in case of f is a surjective function.

asked Oct 14 '18 at 00:16

Fouzi TAKELAIT

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Welcome to MSE, and your solution looks correct! – Xiao Oct 14 '18 at 00:41
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I would say: the intuition is correct, but the proof may or may not be acceptable, depending on the level of rigor expected. – Sort of Damocles Oct 14 '18 at 01:48
See https://math.stackexchange.com/a/989059/589 – lhf Oct 14 '18 at 04:21

Prove, if is surjective function, then is injective function.

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