While the proposed fix of requiring "either that A be inhabited or that B be uninhabited" works, it seems tacked on just to solve this particular edge-case.
I think a more elegant solution would be to soften the definition of a left inverse from a function `g: B -> A` to a function `g: f(A) -> A` where `f(A)` is the subset of elements in `B`, that actually get mapped to by `f` or in the words of the book's function definition, the set of "right" elements in `f`.
This solves the edge-case too, as `f(A) = f({}) = {}` and there exists (exactly one) function `g: {} -> {}`, which also trivially is a left inverse of `f`.
This one is not just in Dummit and Foote; it's just too easy to miss. I'd guess it appears in half the places that state this result. Fixed it in my own lecture notes a few months ago.
I ran into this same thing formalizing some of my old notes in Lean a few days ago. The tricky thing I suppose is that 0. Injectivity and A non-empty or B empty implies left invertibility, 1. Left invertibility implies injectivity. 2. Surjectivity iff right invertibility, and 3. Surjectivity rules out this corner case, so bijectivity iff invertibility. So this one vacuous case just throws a wrench in what is "supposed" to be true.
It warms my heart every time I see an interactive proof assistant being used to improve rather than simply slow down mathematical thinking.
After years of using the things, I believe not enough focus is given to high-velocity uses of proof assistants for prototyping. They can altogether replace scratch paper for fumbling around with new concepts.
I mean, yes, there are a lot of things that are often omitted in mathematical writing and it's up to the reader to infer them (that's "mathematical maturity"). When textbooks discuss intervals, such as [a,b] for example, should the author specify the interval is nondegenerate/nonempty each time? Degenerate cases are often not the primary interest of the particular area or theorem you're studying. We don't usually care about functions with empty or singleton domains. And, yes, technically, you could say a lot of results are technically false due those degenerate/trivial cases.
It's the same situation here. The post proposes an example of a function with a empty domain A. Some authors do actually specify that the domain should be nonempty in this theorem. Others don't. It's not a huge deal.
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).
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.
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.
How does this f satisfy the first criterion, if A is uninhabited? It feels like this function can't be invoked. Am I thinking too much in terms of types here?
When there are no pairs, its certainly true that the first element of each pair comes from A. Just like if there are no living dinosaurs its true that all living dinosaurs speak English.
I was trying to come up with something to explain why I couldn't see it myself: every element of an empty set of integers is both even and odd. This feels counterintuitive to me, until I flip it around into a question: what is the set of all integers that are both even and odd?
I asked AI to formalize an old important paper in analysis. In the paper there is a sequence of epsilon_n > 0, epsilon_n -> 0. It came back, and said: "I formalized it, it is all good, but the assumption that epsilons > 0 is not used anywhere. Shall we remove it, you a get a stronger result this way?"
After years of using the things, I believe not enough focus is given to high-velocity uses of proof assistants for prototyping. They can altogether replace scratch paper for fumbling around with new concepts.
https://github.com/ityonemo/bpa
Its painfully verbose and explicit but its designed to let you cut down to the structure of the proof with a query language
It's the same situation here. The post proposes an example of a function with a empty domain A. Some authors do actually specify that the domain should be nonempty in this theorem. Others don't. It's not a huge deal.
I believe it doesn't make any difference to any meaningful result. It merely makes it easier to write theorems without specifying exceptions.
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).
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.
In other words, the statement "for every x in {} it holds that <anything>" is always true.
The first criterion for a function is stated as:
> The first item in each pair comes from A.
The counter-evidence for the proposition says:
> Let A = {}, and B = {1}. Let f: A -> B = {}
How does this f satisfy the first criterion, if A is uninhabited? It feels like this function can't be invoked. Am I thinking too much in terms of types here?
I was trying to come up with something to explain why I couldn't see it myself: every element of an empty set of integers is both even and odd. This feels counterintuitive to me, until I flip it around into a question: what is the set of all integers that are both even and odd?
LOL