4 ms·

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 n

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."

troethe 1mo ago
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.
- ndriscoll 1mo ago
  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).
  - troethe 1mo ago
    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`.
    - ndriscoll 1mo ago
      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 1mo ago
        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.

troethe 1mo ago

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.

ndriscoll 1mo ago
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).
- troethe 1mo ago
  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`.
  - ndriscoll 1mo ago
    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 1mo ago
      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.