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by csense 386 days ago
> injections on finite sets are also surjections

Not necessarily [1]. I think you're missing an assumption there.

[1] https://en.wikipedia.org/wiki/Injective_function#/media/File...

2 comments

In this case, multiplication by any nonzero fixed element of the ring is an injection from the ring to itself. Any injection from a finite set to itself is indeed a surjection (and so also a bijection).
The intended point, I believe, is the fact that any injective function from a finite set to itself is also surjective.