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Why does `Either.bifoldMap` take a `Monoid` argument, when the monoid basically does no work, since `bifoldMap` always combines the mapped values of the `Either`  with `empty`, where the result of that combination is know ahead of time to _always_ be the same as the mapped values.  In other words, why jump through the hoop of combining a value with `empty`, when we know the result will always be the same original value?

The only reason I can think of is so that `bifoldMap` can build on top of `bifoldLeft`.  Is that it?

I believe `bifoldMap` was previously part of a typeclass that has since been removed from Arrow, which needed the monoid instance to be general for other "bifoldable" types like tuples. Now that it is specialised to `Either`, I can't see any need for it.

Ah, that makes sense.  So maybe there are future plans to remove that dependency on `Monad` and simplify `bifoldMap`s signature.  Thanks <@UQT3B0JSD>.

Hi <@UJHCLJVUK> <@UQT3B0JSD>, bifoldMap uses the Monoid.combine in addition to `empty` in bifoldmap. it’s kind of hidden because it’s achieved through `run`, but when below we use `combine` , that is syntax that is projected by the monoid receiver. I order to replace this monoid here we would have to provide both an empty operation and associate combine operation. <https://github.com/arrow-kt/arrow/blob/b4714662adcd321cc8a03d01ff86da8fb187f406/arrow-libs/core/arrow-core/src/commonMain/kotlin/arrow/core/Either.kt#L822|https://github.com/arrow-kt/arrow/blob/b4714662adcd321cc8a03d01ff86da8fb187f406/ar[…]libs/core/arrow-core/src/commonMain/kotlin/arrow/core/Either.kt>

~I think it’s correct as it’s stated now~

Thanks <@U4UGS5FC7>.  But why use `combine` at all?  Since we know ahead of time that for any type `C`, `MN.empty().combine(f(a))` == `f(a)`.

yes makes sense, it should be doable just with the empty identity, it was probably like that because of the type classes as <@UQT3B0JSD> noted

We should probably review all methods like that to see if there is others. I guess the combine here would only matter if the monoid is unlawful and empty is truly not an empty identity, otherwise as you are saying is essentially a no op in both branches