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 agree! Martín Escardó never tires to say that he uses the Agda proof assistant in exactly this sense, as a kind of interactive blackboard for taking notes and structuring his thoughts. The vast TypeTopology repository is the result of years of following this philosophy: https://github.com/martinescardo/TypeTopology
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`.
The real problem here was, that the statement `g: B -> A` needlessly required `g` to map back elements in B to A, that couldn't even be produced by `f` and should therefore be irrelevant for a left inverse.
That's basically saying you'll just take all functions to be surjective though, and it's stronger than you really need; the non-surjective case works fine for non-empty A.
You could of course interpret some of these basic theorems as saying "well I'd might as well take my function to be surjective since the 'meat' is that case." Much like you could just take all functions to be injective by modding out the kernel since that's the real "meat." And indeed one might interpret the first isomorphism theorem as saying exactly those two things: the isomorphism A/ker f = im f is "the real substance of the map f."
I think a slightly better fix is to change definitions to allow g = { (1, {}) } to be regarded as a left-inverse to g, that is, to allow left-inverses to be partial functions, rather than full functions. The definition still requires they be defined on the image of f, but no choices have to be made on the complement of the image. Probably this breaks some other definitions but it seems intuitively correct to me. It keeps the structure that function B->A could be a left inverse and then only some of them are, rather than limiting them to the functions which are defined only on image(f).
This is kinda nice also because it means that for e.g. the function (a,b) -> (1, 2) given by f(a) = 1, f(b) = 1, you don't need its left inverse to specify that g(2) = a or b, but instead you can have g(2) = {} which doesn't require making any non-canonical choices.
(I'm too sleepy atm to think through this in detail. I might regret this proposal after a nap)
> your fix kinda breaks a lot of the structure of algebra in other ways
What are you referring to here in particular?
I think the property of `g` to be a well defined function is a lot more important than for its domain to be `B`, when `f(A)` is enough to make the composition well defined.
What I would add here is that the property of left-cancellation is exactly equivalent to injectivity, i.e., f : A -> B is injective iff, for any g, h : C -> A, f o g = f o h implies g = h. If A = {} then f is injective and left-cancellative, both vacuously.
The subtlety is now that left-cancellativity is not equivalent to having a left inverse, for exactly the reason pointed out.
The value of this observation is that left-cancellativity is a useful generalization of injectivity that works in any category, where left-cancellative morphisms are called monomorphisms. If you already know about monomorphisms, it's easier to notice that there's something "off" about D&F's exercise!
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 (the empty function is vacuously bijective and its own inverse).
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?"
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 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? That is, should we repeatedly see "a<b" as part of the hypothesis? 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, you could say a lot of results are technically false due to those degenerate/trivial cases. But usually it just means the author didn't want to clutter their writing, or it's not significant to the rest of the theory.
The post proposes a counterexample of a function with a empty domain A. Some authors do actually specify that the domain should be nonempty in this theorem. This is a common result. Others authors don't. It's not a huge deal.
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.
WIP, but that is the target ethos in the prover I'm building:
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
I agree! Martín Escardó never tires to say that he uses the Agda proof assistant in exactly this sense, as a kind of interactive blackboard for taking notes and structuring his thoughts. The vast TypeTopology repository is the result of years of following this philosophy: https://github.com/martinescardo/TypeTopology
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`.
The real problem here was, that the statement `g: B -> A` needlessly required `g` to map back elements in B to A, that couldn't even be produced by `f` and should therefore be irrelevant for a left inverse.
That's basically saying you'll just take all functions to be surjective though, and it's stronger than you really need; the non-surjective case works fine for non-empty A.
You could of course interpret some of these basic theorems as saying "well I'd might as well take my function to be surjective since the 'meat' is that case." Much like you could just take all functions to be injective by modding out the kernel since that's the real "meat." And indeed one might interpret the first isomorphism theorem as saying exactly those two things: the isomorphism A/ker f = im f is "the real substance of the map f."
No, f can still map to `B` and does not need to be surjective. We just loosened the definition of `g` a little in a way that doesn't matter.
I think a slightly better fix is to change definitions to allow g = { (1, {}) } to be regarded as a left-inverse to g, that is, to allow left-inverses to be partial functions, rather than full functions. The definition still requires they be defined on the image of f, but no choices have to be made on the complement of the image. Probably this breaks some other definitions but it seems intuitively correct to me. It keeps the structure that function B->A could be a left inverse and then only some of them are, rather than limiting them to the functions which are defined only on image(f).
This is kinda nice also because it means that for e.g. the function (a,b) -> (1, 2) given by f(a) = 1, f(b) = 1, you don't need its left inverse to specify that g(2) = a or b, but instead you can have g(2) = {} which doesn't require making any non-canonical choices.
(I'm too sleepy atm to think through this in detail. I might regret this proposal after a nap)
> your fix kinda breaks a lot of the structure of algebra in other ways
What are you referring to here in particular?
I think the property of `g` to be a well defined function is a lot more important than for its domain to be `B`, when `f(A)` is enough to make the composition well defined.
What I would add here is that the property of left-cancellation is exactly equivalent to injectivity, i.e., f : A -> B is injective iff, for any g, h : C -> A, f o g = f o h implies g = h. If A = {} then f is injective and left-cancellative, both vacuously.
The subtlety is now that left-cancellativity is not equivalent to having a left inverse, for exactly the reason pointed out.
The value of this observation is that left-cancellativity is a useful generalization of injectivity that works in any category, where left-cancellative morphisms are called monomorphisms. If you already know about monomorphisms, it's easier to notice that there's something "off" about D&F's exercise!
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.
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 (the empty function is vacuously bijective and its own inverse).
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.
What can be confusing is that the statement "for every x in {}, it doesn’t hold that <anything>" is always true as well.
But that is the definition of injective.
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?"
LOL
Was the proof correct?
No you see people that use AI generally don't bother to consider this unimportant detail. Or they ask the AI to "double check" its work.
Help me out, I feel dumb.
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?
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.
That helps. Thank you!
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 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? That is, should we repeatedly see "a<b" as part of the hypothesis? 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, you could say a lot of results are technically false due to those degenerate/trivial cases. But usually it just means the author didn't want to clutter their writing, or it's not significant to the rest of the theory.
The post proposes a counterexample of a function with a empty domain A. Some authors do actually specify that the domain should be nonempty in this theorem. This is a common result. Others authors don't. It's not a huge deal.