`φ`

and `γ`

be members of `T`, that is they are natural transformations from `F` to `MF'` and from `G` to `MG'`, respectively. Let them be such that `F' = G`. Now `(M γ)` will also be a natural transformation, formed by composing the functor `M` with the natural transformation `γ`

. Similarly, `(join G')` will be a natural transformation, formed by composing the natural transformation `join` with the functor `G'`; it will transform the functor `MMG'` to the functor `MG'`. Now take the vertical composition of the three natural transformations `(join G')`, `(M γ)`

, and `φ`

, and abbreviate it as follows:
+Let `φ`

and `γ`

be members of `T`, that is they are natural transformations from `F` to `MF'` and from `G` to `MG'`, respectively. Let them be such that `F' = G`. Now `(M γ)`

will also be a natural transformation, formed by composing the functor `M` with the natural transformation `γ`

. Similarly, `(join G')` will be a natural transformation, formed by composing the natural transformation `join` with the functor `G'`; it will transform the functor `MMG'` to the functor `MG'`. Now take the vertical composition of the three natural transformations `(join G')`, `(M γ)`

, and `φ`

, and abbreviate it as follows:
γ <=< φ =def. ((join G') -v- (M γ) -v- φ) @@ -204,20 +204,20 @@ Since composition is associative I don't specify the order of composition on the In other words, `<=<` is a binary operator that takes us from two members`φ`

and`γ`

of `T` to a composite natural transformation. (In functional programming, at least, this is called the "Kleisli composition operator". Sometimes its written`φ >=> γ`

where that's the same as`γ <=< φ`

.) -φ is a transformation from `F` to `MF'` which = `MG`; `(M γ)` is a transformation from `MG` to `MMG'`; and `(join G')` is a transformation from `MMG'` to `MG'`. So the composite γ <=< φ will be a transformation from `F` to `MG'`, and so also eligible to be a member of `T`. +`φ`

is a transformation from `F` to `MF'` which = `MG`;`(M γ)`

is a transformation from `MG` to `MMG'`; and `(join G')` is a transformation from `MMG'` to `MG'`. So the composite`γ <=< φ`

will be a transformation from `F` to `MG'`, and so also eligible to be a member of `T`. Now we can specify the "monad laws" governing a monad as follows: (T, <=<, unit) constitute a monoid -That's it. (Well, perhaps we're cheating a bit, because γ <=< φ isn't fully defined on `T`, but only when `F` is a functor to `MF'` and `G` is a functor from `F'`. But wherever `<=<` is defined, the monoid laws are satisfied: +That's it. Well, there may be a wrinkle here. I don't know whether the definition of a monoid requires the operation to be defined for every pair in its set. In the present case,`γ <=< φ`

isn't fully defined on `T`, but only when `F` is a functor to `MF'` and `G` is a functor from `F'`. But wherever `<=<` is defined, the monoid laws are satisfied: - (i) γ <=< φ is also in T - (ii) (ρ <=< γ) <=< φ = ρ <=< (γ <=< φ) + (i) γ <=< φ is also in T + (ii) (ρ <=< γ) <=< φ = ρ <=< (γ <=< φ) (iii.1) unit <=< φ = φ (here φ has to be a natural transformation to M(1C)) (iii.2) φ = φ <=< unit (here φ has to be a natural transformation from 1C) -If φ is a natural transformation from `F` to `M(1C)` and γ is `(φ G')`, that is, a natural transformation from `PG` to `MG`, then we can extend (iii.1) as follows: +If`φ`

is a natural transformation from `F` to `M(1C)` and`γ`

is`(φ G')`

, that is, a natural transformation from `FG` to `MG`, then we can extend (iii.1) as follows: γ = (φ G') = ((unit <=< φ) G') @@ -227,7 +227,7 @@ If φ is a natural transformation from `F` to `M(1C)` and γ is `(φ ?? = (unit G') <=< γ -where as we said γ is a natural transformation from some `PG'` to `MG'`. +where as we said`γ`

is a natural transformation from some `FG'` to `MG'`. Similarly, if φ is a natural transformation from `1C` to `MF'`, and γ is `(φ G)`, that is, a natural transformation from `G` to `MF'G`, then we can extend (iii.2) as follows: -- 2.11.0