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.

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ndriscoll 6 hours ago | parent | next [-]

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."

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troethe 5 hours ago | parent [-]

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.

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ndriscoll 5 hours ago | parent [-]

But f's codomain is B, and g isn't a function on B, so you can't compose them in the first place. And saying "well yeah but you could compose f's restriction" is exactly making f surjective.

The basic result here is every function factors as a surjection (collapsing to the quotient) followed by an isomorphism (with the image) followed by an injection (enlarging the codomain). The surjection and injection are "trivial" and the isomorphism is the part that "does something" (permuting your thing somehow).

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troethe 5 hours ago | parent [-]

Of course I can compose `f: A -> B` and `g: f(A) -> A`. The composition maps a `x` from `A` to `g(f(x))` which is well defined. Therefore the composition is a function.

`g` and `f` aren't functions in a programming language and `A` and `B` are not types. There is nothing like a type checker forbidding you from composing `f: A -> B` and `g: f(A) -> A`.

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ndriscoll 5 hours ago | parent [-]

I suppose, but then your definition of composition requires checking compatibility of codomain/domain instead of it being automatic (and your definition of composition does not fit the shape to form a category), which seems way worse than stipulating f non-empty for some theorem that isn't even true for other categories either.

And for e.g. for vector spaces or modules over semisimple rings they're actually the same so there's a real statement being made to say they're not always.

	▲	troethe 4 hours ago \| parent [-]
		I think we can sit here and speculate which properties of left inverses are actually important for quite a while. But we will only ever get a real answer to this question, when the original poster finishes formalizing the whole book and can tell us which weakenings break further proofs in the books and which won't.

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ajkjk 5 hours ago | parent | prev [-]

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)

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troethe 5 hours ago | parent [-]

> 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.

	▲	ndriscoll 5 hours ago \| parent \| next [-]
		Typically one defines relations before functions anyway (unless you're doing type theory/programming, in which case types matter), and relations also offer a fix (the empty relation from the empty set is left inverted by the empty relation from the target to the empty set). Most algebra books I've read are either explicitly or at least implicitly setting up structural analogies to introduce categories, where your A and B are indeed fixed/"typed".
	▲	ajkjk 5 hours ago \| parent \| prev [-]
		sorry I edited my comment because I wasn't quite sure what I meant by that either. I am too sleepy. But, I disagree. partial functions B->A are just a superset of functions B->A, so requiring g to be partial is just relaxing a requirement. Whereas requiring g: im(A)->A adds an additional restriction, potentially breaking any place that concept of a left inverse is used. Invariably someone will just define a new thing equivalent to what is normally called the left inverse because they still need the concept for some reason.