Timeline for If we don't use function composition, does Maybe remain a monad?
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7 events
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| Jul 18, 2022 at 12:39 | comment | added | pro100tom | I've took some time to digest what you have said and although your answer does help, I am still not convinced (I disagree) with your second point. Function composition is a binay operation and a monoid must have that binary operation (otherwise it's not a monoid). Since monad is a monoid and function composition is a binary operation, then it is in fact the part of the definition of the monad. I agree with the "if you don't use it it doesn't affect the fact that it is a monad". But I think that function composition must be defined somewhere. Hence, my question remains open. | |
| Jul 18, 2022 at 12:30 | comment | added | Robert Bräutigam | Yes, I mean identity element. | |
| Jul 18, 2022 at 11:52 | comment | added | pro100tom | When you write "null element" do you mean the "identity element"? | |
| Jul 18, 2022 at 8:11 | comment | added | Robert Bräutigam | Regarding 3rd point: You're right, that we sometimes call some object functor, sometimes a data structure and we don't even always specify the function to it, etc. In the end it doesn't matter, it only matters whether the requirements of the definition are met. Can you construct the tuple: set, null element, function with the necessary properties? If yes, it's a functor. It doesn't matter what syntax it is in. Does that help? | |
| Jul 17, 2022 at 20:01 | vote | accept | pro100tom | ||
| Jul 18, 2022 at 12:33 | |||||
| Jul 17, 2022 at 19:12 | comment | added | pro100tom | Thank you for your response. It makes sense! However, can you please elaborate more on the 3rd point please? I didn't get that very explanation about the objects being monads instead of functions. | |
| Jul 17, 2022 at 18:41 | history | answered | Robert Bräutigam | CC BY-SA 4.0 |