I don’t think it’s fair to call {}-> injective just because no two inputs map to the same output. That’s vacuous.

Generally mathematicians treat vacuous statements as true.

I believe it doesn't make any difference to any meaningful result. It merely makes it easier to write theorems without specifying exceptions.

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gpm 6 hours ago | parent | prev | next [-]

Edit: Removed incorrect claim that |B| > |A| sufficed for the counter example.

It's also the definitions the book supplies though (and the standard ones). Mathematics works over definitions. Everyone is free to do math over whatever definitions they want - but what is or isn't true follows from them. Lots of definitions and theorems exclude things like empty-set cases because they're weird, but that has to be explicit (otherwise someone will apply a theorem to the empty set and it will lead them to incorrect conclusions).

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ndriscoll 6 hours ago | parent [-]

No, empty A is critical to the counterexample. In your example, g(x) = 1 is a left inverse.

The point is you either send an element of the codomain to its (unique by injectivity) preimage if it's in the image, or to an arbitrary element of A if it's not, and that's a left inverse. But then if B has an element, A needs one for you to pick your arbitrary target.

In a sense, your claim that the problem is a smaller domain than codomain does contribute though; if f is also surjective, then this case can't happen, so bijective iff invertible (the empty function is vacuously bijective and its own inverse).

	▲	gpm 6 hours ago \| parent [-]
		Oh, oops, you're right. Sorry.

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tim-kt 6 hours ago | parent | prev | next [-]

It's true precisely because it's vacuous. If you quantify over the empty set, anything is true.

In other words, the statement "for every x in {} it holds that <anything>" is always true.

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layer8 6 hours ago | parent [-]

What can be confusing is that the statement "for every x in {}, it doesn’t hold that <anything>" is always true as well.

	▲	tim-kt 6 hours ago \| parent [-]
		I mean, yes. But "it doesn't hold that <anything>" is equivalent to "it holds that <not anything>" and since not anything is also anything... Ah, I see.

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mitxela 5 hours ago | parent | prev [-]

But that is the definition of injective.