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Mathematicans like to call two things with the same "structure" by the same name, even if it's not obvious how they're otherwise related. In this case, the shar

Mathematicans like to call two things with the same "structure" by the same name, even if it's not obvious how they're otherwise related. In this case, the shared structure is that both the derivative of a function and D(n) follow the product rule of derivatives. Sharing a structure means two mathematical objects are "morphic" (homomorphic[1], isomorphic[2], etc.) in some way, so in some sense they are the are same object. In this case, we can construct a morphism. Since D(n) follows the product rule, we only need to find a function f of x for each prime p which at some x has a value of p and a derivative of 1. Then we can compose those functions by multiplication for all other natural numbers. f_p(x) = x + p is one such set of functions, giving us the complete function F_n(x) = Π_(p∈P(n)) x + p, where P(n) is the set of prime factors of n. Note, the product of the empty set is defined to be 1, so F_1(x) = 1. Finally, the homomorphism between D and the derivatives of functions is that D(n) = F'_n(0), so in some sense, D really is a derivative. [1]: https://en.wikipedia.org/wiki/Homomorphism https://en.wikipedia.org/wiki/Homomorphism [2]: https://en.wikipedia.org/wiki/Isomorphism https://en.wikipedia.org/wiki/Isomorphism

supernikio2 2y ago
I'm rather new to the concept of objects and morphisms myself, but I love how fields like category theory allow one to "zoom out" far enough in abstraction that two seemingly different concepts are actually the same applied to distinct contexts.
BalinKing 2y ago
> so in some sense they are the are same object. My understanding is that this is only the case if they’re isomorphic—even a pair of homomorphisms between the objects is not itself sufficient to identify them as the same. But I also don’t know any category theory, so I might be spouting nonsense :-)
- GrantMoyer 2y ago
  Indeed, a homomorphism from A to B is an isomorphism from A to a subset of B. In other words, homomorphism is to injective as isomorphism is to bijective. So whether you say A is equivalent to B depends on how you define equivalent. For anyone curious, while Category Theory is very concerned with morphisms, they also come up in many other places. In particular, Abstract Algebra (groups and rings and such) may be a more approachable introduction to morphisms than Category Theory, and then Category Theory flows pretty naturally from the concepts of Abstract Algebra.
  - smohare 2y ago
    A generic homomorphism does not induce any isomorphism to a subset of B that I can think of unless it is already injective. The isomorphism you’re after is rather from the quotient structure of A mod the kernel to the image.
    - alexanderchr 2y ago
      Simply consider any constant morphism and it is obvious that OPs statement is false. As you say one needs injectivity.

supernikio2 2y ago

I'm rather new to the concept of objects and morphisms myself, but I love how fields like category theory allow one to "zoom out" far enough in abstraction that two seemingly different concepts are actually the same applied to distinct contexts.

BalinKing 2y ago

> so in some sense they are the are same object. My understanding is that this is only the case if they’re isomorphic—even a pair of homomorphisms between the objects is not itself sufficient to identify them as the same. But I also don’t know any category theory, so I might be spouting nonsense :-)

GrantMoyer 2y ago
Indeed, a homomorphism from A to B is an isomorphism from A to a subset of B. In other words, homomorphism is to injective as isomorphism is to bijective. So whether you say A is equivalent to B depends on how you define equivalent. For anyone curious, while Category Theory is very concerned with morphisms, they also come up in many other places. In particular, Abstract Algebra (groups and rings and such) may be a more approachable introduction to morphisms than Category Theory, and then Category Theory flows pretty naturally from the concepts of Abstract Algebra.
- smohare 2y ago
  A generic homomorphism does not induce any isomorphism to a subset of B that I can think of unless it is already injective. The isomorphism you’re after is rather from the quotient structure of A mod the kernel to the image.
  - alexanderchr 2y ago
    Simply consider any constant morphism and it is obvious that OPs statement is false. As you say one needs injectivity.