Thank you for the helpful response and pointers. I think I need to take my time to sort this out, thanks a lot and I'll come back with a better understanding, hopefully!
I did my homework. Now I'm not so certain about everything, so please take this with a grain of salt.
Typescript's generic functions seem to act like they're contravariant over the type parameter's type bound. I'm not entirely sure though.
// Assignment succeeds with F<unknown> -> F<string>
// Assignment fails with F<string> -> F<unknown>
const a1: <T extends unknown>(x: T) => T = (x) => x;
const b1: <U extends string>(y: U) => U = a1;
const c1: typeof a1 = b1; // error
const a2: <T extends unknown>(x: T) => string = String;
const b2: <U extends string>(x: U) => string = a2;
const c2: typeof a2 = b2; // error
declare const a3: <T extends unknown>() => T;
const b3: <U extends string>() => U = a3;
const c3: typeof a3 = b3; // error@masaeedu This seems like the {co,contra}variance of the types is independent to whether you can do a {co,contra}map on it. May I ask, did I mess up again, or is this another unsoundness, or is this the way it is? Anyhow thank you so much for the helpful discussion! heart
@masaeedu When you have time, could you take a look at my last post?Edit: this oneEdit: I tried to wait the variance checking issue to resolve and wait for your feedback before posting to the generic values issue, but this snippet keeps popping up in my head, several times a day. I needed to dump this snippet in the generic values issue and move on with my life. I hope you understand...