* finite strings of an alphabet `A`, with <code>⋆</code> being concatenation and `z` being the empty string
* all functions <code>X→X</code> over a set `X`, with <code>⋆</code> being composition and `z` being the identity function over `X`
-* the natural numbers with <code>⋆</code> being plus and `z` being `0` (in particular, this is a **commutative monoid**). If we use the integers, or the naturals mod n, instead of the naturals, then every element will have an inverse and so we have not merely a monoid but a **group**.)
-* if we let <code>⋆</code> be multiplication and `z` be `1`, we get different monoids over the same sets as in the previous item.
+* the natural numbers with <code>⋆</code> being plus and `z` being 0 (in particular, this is a **commutative monoid**). If we use the integers, or the naturals mod n, instead of the naturals, then every element will have an inverse and so we have not merely a monoid but a **group**.
+* if we let <code>⋆</code> be multiplication and `z` be 1, we get different monoids over the same sets as in the previous item.
Categories
----------
These parallel the constraints for monoids. Note that there can be multiple distinct morphisms between an element `E` and itself; they need not all be identity morphisms. Indeed from (iii) it follows that each element can have only a single identity morphism.
-A good intuitive picture of a category is as a generalized directed graph, where the category elements are the graph's nodes, and there can be multiple directed edges between a given pair of nodes, and nodes can also have multiple directed edges to themselves. (Every node must have at least one such, which is that node's identity morphism.)
+A good intuitive picture of a category is as a generalized directed graph, where the category elements are the graph's nodes, and there can be multiple directed edges between a given pair of nodes, and nodes can also have multiple directed edges to themselves. Morphisms correspond to directed paths of length ≥ 0 in the graph.
Some examples of categories are:
-* Categories whose elements are sets and whose morphisms are functions between those sets. Here the source and target of a function are its domain and range, so distinct functions sharing a domain and range (e.g., sin and cos) are distinct morphisms between the same source and target elements. The identity morphism for any element/set is just the identity function for that set.
+* Categories whose elements are sets and whose morphisms are functions between those sets. Here the source and target of a function are its domain and range, so distinct functions sharing a domain and range (e.g., `sin` and `cos`) are distinct morphisms between the same source and target elements. The identity morphism for any element/set is just the identity function for that set.
* any monoid <code>(S,⋆,z)</code> generates a category with a single element `x`; this `x` need not have any relation to `S`. The members of `S` play the role of *morphisms* of this category, rather than its elements. All of these morphisms are understood to map `x` to itself. The result of composing the morphism consisting of `s1` with the morphism `s2` is the morphism `s3`, where <code>s3=s1⋆s2</code>. The identity morphism for the (single) category element `x` is the monoid's identity `z`.
-* a **preorder** is a structure `(S, ≤)` consisting of a reflexive, transitive, binary relation on a set `S`. It need not be connected (that is, there may be members `x`,`y` of `S` such that neither `x≤y` nor `y≤x`). It need not be anti-symmetric (that is, there may be members `s1`,`s2` of `S` such that `s1≤s2` and `s2≤s1` but `s1` and `s2` are not identical). Some examples:
+* a **preorder** is a structure <code>(S, ≤)</code> consisting of a reflexive, transitive, binary relation on a set `S`. It need not be connected (that is, there may be members `s1`,`s2` of `S` such that neither <code>s1≤s2</code> nor <code>s2≤s1</code>). It need not be anti-symmetric (that is, there may be members `s1`,`s2` of `S` such that <code>s1≤s2</code> and <code>s2≤s1</code> but `s1` and `s2` are not identical). Some examples:
* sentences ordered by logical implication ("p and p" implies and is implied by "p", but these sentences are not identical; so this illustrates a pre-order without anti-symmetry)
* sets ordered by size (this illustrates it too)
A **functor** is a "homomorphism", that is, a structure-preserving mapping, between categories. In particular, a functor `F` from category <b>C</b> to category <b>D</b> must:
<pre>
- (i) associate with every element C1 of <b>C</b> an element F(C1) of <b>D</b>
- (ii) associate with every morphism f:C1→C2 of <b>C</b> a morphism F(f):F(C1)→F(C2) of <b>D</b>
- (iii) "preserve identity", that is, for every element C1 of <b>C</b>: F of C1's identity morphism in <b>C</b> must be the identity morphism of F(C1) in <b>D</b>: F(1<sub>C1</sub>) = 1<sub>F(C1)</sub>.
- (iv) "distribute over composition", that is for any morphisms f and g in <b>C</b>: F(g ∘ f) = F(g) ∘ F(f)
+ (i) associate with every element C1 of <b>C</b> an element F(C1) of <b>D</b>
+
+ (ii) associate with every morphism f:C1→C2 of <b>C</b> a morphism F(f):F(C1)→F(C2) of <b>D</b>
+
+ (iii) "preserve identity", that is, for every element C1 of <b>C</b>:
+ F of C1's identity morphism in <b>C</b> must be the identity morphism of F(C1) in <b>D</b>:
+ F(1<sub>C1</sub>) = 1<sub>F(C1)</sub>.
+
+ (iv) "distribute over composition", that is for any morphisms f and g in <b>C</b>:
+ F(g ∘ f) = F(g) ∘ F(f)
</pre>
A functor that maps a category to itself is called an **endofunctor**. The (endo)functor that maps every element and morphism of <b>C</b> to itself is denoted `1C`.
----------------------
So categories include elements and morphisms. Functors consist of mappings from the elements and morphisms of one category to those of another (or the same) category. **Natural transformations** are a third level of mappings, from one functor to another.
-Where `G` and `H` are functors from category <b>C</b> to category <b>D</b>, a natural transformation η between `G` and `H` is a family of morphisms η[C1]:G(C1)→H(C1)` in <b>D</b> for each element `C1` of <b>C</b>. That is, η[C1]` has as source `C1`'s image under `G` in <b>D</b>, and as target `C1`'s image under `H` in <b>D</b>. The morphisms in this family must also satisfy the constraint:
+Where `G` and `H` are functors from category <b>C</b> to category <b>D</b>, a natural transformation η between `G` and `H` is a family of morphisms <code>η[C1]:G(C1)→H(C1)</code> in <b>D</b> for each element `C1` of <b>C</b>. That is, <code>η[C1]</code> has as source `C1`'s image under `G` in <b>D</b>, and as target `C1`'s image under `H` in <b>D</b>. The morphisms in this family must also satisfy the constraint:
- for every morphism f:C1→C2 in <b>C</b>: η[C2] ∘ G(f) = H(f) ∘ η[C1]
+<pre>
+ for every morphism f:C1→C2 in <b>C</b>:
+ η[C2] ∘ G(f) = H(f) ∘ η[C1]
+</pre>
-That is, the morphism via `G(f)` from `G(C1)` to `G(C2)`, and then via η[C2]` to `H(C2)`, is identical to the morphism from `G(C1)` via η[C1]` to `H(C1)`, and then via `H(f)` from `H(C1)` to `H(C2)`.
+That is, the morphism via `G(f)` from `G(C1)` to `G(C2)`, and then via <code>η[C2]</code> to `H(C2)`, is identical to the morphism from `G(C1)` via <code>η[C1]</code> to `H(C1)`, and then via `H(f)` from `H(C1)` to `H(C2)`.
How natural transformations compose:
Consider four categories <b>B</b>, <b>C</b>, <b>D</b>, and <b>E</b>. Let `F` be a functor from <b>B</b> to <b>C</b>; `G`, `H`, and `J` be functors from <b>C</b> to <b>D</b>; and `K` and `L` be functors from <b>D</b> to <b>E</b>. Let η be a natural transformation from `G` to `H`; φ be a natural transformation from `H` to `J`; and ψ be a natural transformation from `K` to `L`. Pictorally:
+<pre>
- <b>B</b> -+ +--- <b>C</b> --+ +---- <b>D</b> -----+ +-- <b>E</b> --
| | | | | |
- F: -----→ G: -----→ K: -----→
- | | | | | η | | | ψ
+ F: ------> G: ------> K: ------>
+ | | | | | η | | | ψ
| | | | v | | v
- | | H: -----→ L: -----→
- | | | | | φ | |
+ | | H: ------> L: ------>
+ | | | | | φ | |
| | | | v | |
- | | J: -----→ | |
+ | | J: ------> | |
-----+ +--------+ +------------+ +-------
+</pre>
-Then `(η F)` is a natural transformation from the (composite) functor `GF` to the composite functor `HF`, such that where `b1` is an element of category <b>B</b>, `(η F)[b1] = η[F(b1)]`---that is, the morphism in <b>D</b> that η assigns to the element `F(b1)` of <b>C</b>.
+Then <code>(η F)</code> is a natural transformation from the (composite) functor `GF` to the composite functor `HF`, such that where `B1` is an element of category <b>B</b>, <code>(η F)[B1] = η[F(B1)]</code>---that is, the morphism in <b>D</b> that <code>η</code> assigns to the element `F(B1)` of <b>C</b>.
-And `(K η)` is a natural transformation from the (composite) functor `KG` to the (composite) functor `KH`, such that where `C1` is an element of category <b>C</b>, `(K η)[C1] = K(η[C1])`---that is, the morphism in <b>E</b> that `K` assigns to the morphism η[C1]` of <b>D</b>.
+And <code>(K η)</code> is a natural transformation from the (composite) functor `KG` to the (composite) functor `KH`, such that where `C1` is an element of category <b>C</b>, <code>(K η)[C1] = K(η[C1])</code>---that is, the morphism in <b>E</b> that `K` assigns to the morphism <code>η[C1]</code> of <b>D</b>.
-`(φ -v- η)` is a natural transformation from `G` to `J`; this is known as a "vertical composition". We will rely later on this, where `f:C1→C2`:
+<code>(φ -v- η)</code> is a natural transformation from `G` to `J`; this is known as a "vertical composition". For any morphism <code>f:C1→C2</code> in <b>C</b>:
+<pre>
φ[C2] ∘ H(f) ∘ η[C1] = φ[C2] ∘ H(f) ∘ η[C1]
+</pre>
-by naturalness of φ, is:
+by naturalness of <code>φ</code>, is:
+<pre>
φ[C2] ∘ H(f) ∘ η[C1] = J(f) ∘ φ[C1] ∘ η[C1]
+</pre>
-by naturalness of η, is:
+by naturalness of <code>η</code>, is:
+<pre>
φ[C2] ∘ η[C2] ∘ G(f) = J(f) ∘ φ[C1] ∘ η[C1]
+</pre>
-Hence, we can define `(φ -v- η)[x]` as: φ[x] ∘ η[x]` and rely on it to satisfy the constraints for a natural transformation from `G` to `J`:
+Hence, we can define <code>(φ -v- η)[\_]</code> as: <code>φ[\_] ∘ η[\_]</code> and rely on it to satisfy the constraints for a natural transformation from `G` to `J`:
+<pre>
(φ -v- η)[C2] ∘ G(f) = J(f) ∘ (φ -v- η)[C1]
+</pre>
An observation we'll rely on later: given the definitions of vertical composition and of how natural transformations compose with functors, it follows that:
+<pre>
((φ -v- η) F) = ((φ F) -v- (η F))
+</pre>
I'll assert without proving that vertical composition is associative and has an identity, which we'll call "the identity transformation."
-`(ψ -h- η)` is natural transformation from the (composite) functor `KG` to the (composite) functor `LH`; this is known as a "horizontal composition." It's trickier to define, but we won't be using it here. For reference:
+<code>(ψ -h- η)</code> is natural transformation from the (composite) functor `KG` to the (composite) functor `LH`; this is known as a "horizontal composition." It's trickier to define, but we won't be using it here. For reference:
+<pre>
(φ -h- η)[C1] = L(η[C1]) ∘ ψ[G(C1)]
- = ψ[H(C1)] ∘ K(η[C1])
+ = ψ[H(C1)] ∘ K(η[C1])
+</pre>
Horizontal composition is also associative, and has the same identity as vertical composition.
A **monad** is a structure consisting of an (endo)functor `M` from some category <b>C</b> to itself, along with some natural transformations, which we'll specify in a moment.
-Let `T` be a set of natural transformations `p`, each being between some (variable) functor `P` and another functor which is the composite `MP'` of `M` and a (variable) functor `P'`. That is, for each element `C1` in <b>C</b>, `p` assigns `C1` a morphism from element `P(C1)` to element `MP'(C1)`, satisfying the constraints detailed in the previous section. For different members of `T`, the relevant functors may differ; that is, `p` is a transformation from functor `P` to `MP'`, `q` is a transformation from functor `Q` to `MQ'`, and none of `P`, `P'`, `Q`, `Q'` need be the same.
+Let `T` be a set of natural transformations <code>φ</code>, each being between some arbitrary endofunctor `F` on <b>C</b> and another functor which is the composite `MF'` of `M` and another arbitrary endofunctor `F'` on <b>C</b>. That is, for each element `C1` in <b>C</b>, <code>φ</code> assigns `C1` a morphism from element `F(C1)` to element `MF'(C1)`, satisfying the constraints detailed in the previous section. For different members of `T`, the relevant functors may differ; that is, <code>φ</code> is a transformation from functor `F` to `MF'`, <code>γ</code> is a transformation from functor `G` to `MG'`, and none of `F`, `F'`, `G`, `G'` need be the same.
-One of the members of `T` will be designated the "unit" transformation for `M`, and it will be a transformation from the identity functor `1C` for <b>C</b> to `M(1C)`. So it will assign to `C1` a morphism from `C1` to `M(C1)`.
+One of the members of `T` will be designated the `unit` transformation for `M`, and it will be a transformation from the identity functor `1C` for <b>C</b> to `M(1C)`. So it will assign to `C1` a morphism from `C1` to `M(C1)`.
-We also need to designate for `M` a "join" transformation, which is a natural transformation from the (composite) functor `MM` to `M`.
+We also need to designate for `M` a `join` transformation, which is a natural transformation from the (composite) functor `MM` to `M`.
These two natural transformations have to satisfy some constraints ("the monad laws") which are most easily stated if we can introduce a defined notion.
-Let `p` and `q` be members of `T`, that is they are natural transformations from `P` to `MP'` and from `Q` to `MQ'`, respectively. Let them be such that `P' = Q`. Now `(M q)` will also be a natural transformation, formed by composing the functor `M` with the natural transformation `q`. Similarly, `(join Q')` will be a natural transformation, formed by composing the natural transformation `join` with the functor `Q'`; it will transform the functor `MMQ'` to the functor `MQ'`. Now take the vertical composition of the three natural transformations `(join Q')`, `(M q)`, and `p`, and abbreviate it as follows:
+Let <code>φ</code> and <code>γ</code> 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 <code>(M γ)</code> will also be a natural transformation, formed by composing the functor `M` with the natural transformation <code>γ</code>. 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')`, <code>(M γ)</code>, and <code>φ</code>, and abbreviate it as follows. Since composition is associative I don't specify the order of composition on the rhs.
- q <=< p =def. ((join Q') -v- (M q) -v- p)
-
-Since composition is associative I don't specify the order of composition on the rhs.
+<pre>
+ γ <=< φ =def. ((join G') -v- (M γ) -v- φ)
+</pre>
-In other words, `<=<` is a binary operator that takes us from two members `p` and `q` of `T` to a composite natural transformation. (In functional programming, at least, this is called the "Kleisli composition operator". Sometimes its written `p >=> q` where that's the same as `q <=< p`.)
+In other words, `<=<` is a binary operator that takes us from two members <code>φ</code> and <code>γ</code> of `T` to a composite natural transformation. (In functional programming, at least, this is called the "Kleisli composition operator". Sometimes it's written <code>φ >=> γ</code> where that's the same as <code>γ <=< φ</code>.)
-`p` is a transformation from `P` to `MP'` which = `MQ`; `(M q)` is a transformation from `MQ` to `MMQ'`; and `(join Q')` is a transformation from `MMQ'` to `MQ'`. So the composite `q <=< p` will be a transformation from `P` to `MQ'`, and so also eligible to be a member of `T`.
+<code>φ</code> is a transformation from `F` to `MF'`, where the latter = `MG`; <code>(M γ)</code> is a transformation from `MG` to `MMG'`; and `(join G')` is a transformation from `MMG'` to `MG'`. So the composite <code>γ <=< φ</code> 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:
+<pre>
(T, <=<, unit) constitute a monoid
+</pre>
+
+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, <code>γ <=< φ</code> isn't fully defined on `T`, but only when <code>φ</code> is a transformation to some `MF'` and <code>γ</code> is a transformation from `F'`. But wherever `<=<` is defined, the monoid laws must hold:
+
+<pre>
+ (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) ρ = ρ <=< unit
+ (here ρ has to be a natural transformation from 1C)
+</pre>
+
+If <code>φ</code> is a natural transformation from `F` to `M(1C)` and <code>γ</code> is <code>(φ G')</code>, that is, a natural transformation from `FG'` to `MG'`, then we can extend (iii.1) as follows:
-That's it. (Well, perhaps we're cheating a bit, because `q <=< p` isn't fully defined on `T`, but only when `P` is a functor to `MP'` and `Q` is a functor from `P'`. But wherever `<=<` is defined, the monoid laws are satisfied:
+<pre>
+ γ = (φ G')
+ = ((unit <=< φ) G')
+ = (((join 1C) -v- (M unit) -v- φ) G')
+ = (((join 1C) G') -v- ((M unit) G') -v- (φ G'))
+ = ((join (1C G')) -v- (M (unit G')) -v- γ)
+ = ((join G') -v- (M (unit G')) -v- γ)
+ since (unit G') is a natural transformation to MG',
+ this satisfies the definition for <=<:
+ = (unit G') <=< γ
+</pre>
+
+where as we said <code>γ</code> is a natural transformation from some `FG'` to `MG'`.
- (i) q <=< p is also in T
- (ii) (r <=< q) <=< p = r <=< (q <=< p)
- (iii.1) unit <=< p = p (here p has to be a natural transformation to M(1C))
- (iii.2) p = p <=< unit (here p has to be a natural transformation from 1C)
+Similarly, if <code>ρ</code> is a natural transformation from `1C` to `MR'`, and <code>γ</code> is <code>(ρ G)</code>, that is, a natural transformation from `G` to `MR'G`, then we can extend (iii.2) as follows:
+
+<pre>
+ γ = (ρ G)
+ = ((ρ <=< unit) G)
+ = (((join R') -v- (M ρ) -v- unit) G)
+ = (((join R') G) -v- ((M ρ) G) -v- (unit G))
+ = ((join (R'G)) -v- (M (ρ G)) -v- (unit G))
+ since γ = (ρ G) is a natural transformation to MR'G,
+ this satisfies the definition <=<:
+ = γ <=< (unit G)
+</pre>
-If `p` is a natural transformation from `P` to `M(1C)` and `q` is `(p Q')`, that is, a natural transformation from `PQ` to `MQ`, then we can extend (iii.1) as follows:
+where as we said <code>γ</code> is a natural transformation from `G` to some `MR'G`.
- q = (p Q')
- = ((unit <=< p) Q')
- = ((join -v- (M unit) -v- p) Q')
- = (join Q') -v- ((M unit) Q') -v- (p Q')
- = (join Q') -v- (M (unit Q')) -v- q
- ??
- = (unit Q') <=< q
+Summarizing then, the monad laws can be expressed as:
-where as we said `q` is a natural transformation from some `PQ'` to `MQ'`.
+<pre>
+ For all ρ, γ, φ in T for which ρ <=< γ and γ <=< φ are defined:
-Similarly, if `p` is a natural transformation from `1C` to `MP'`, and `q` is `(p Q)`, that is, a natural transformation from `Q` to `MP'Q`, then we can extend (iii.2) as follows:
+ (i) γ <=< φ etc are also in T
- q = (p Q)
- = ((p <=< unit) Q)
- = (((join P') -v- (M p) -v- unit) Q)
- = ((join P'Q) -v- ((M p) Q) -v- (unit Q))
- = ((join P'Q) -v- (M (p Q)) -v- (unit Q))
- ??
- = q <=< (unit Q)
+ (ii) (ρ <=< γ) <=< φ = ρ <=< (γ <=< φ)
-where as we said `q` is a natural transformation from `Q` to some `MP'Q`.
+ (iii.1) (unit G') <=< γ = γ
+ when γ is a natural transformation from some FG' to MG'
+ (iii.2) γ = γ <=< (unit G)
+ when γ is a natural transformation from G to some MR'G
+</pre>
-----------------------------------------------------------
In category theory, the monad laws are usually stated in terms of `unit` and `join` instead of `unit` and `<=<`.
-(*
+<!--
P2. every element C1 of a category <b>C</b> has an identity morphism 1<sub>C1</sub> such that for every morphism f:C1→C2 in <b>C</b>: 1<sub>C2</sub> ∘ f = f = f ∘ 1<sub>C1</sub>.
P3. functors "preserve identity", that is for every element C1 in F's source category: F(1<sub>C1</sub>) = 1<sub>F(C1)</sub>.
-*)
+-->
Let's remind ourselves of some principles:
- * composition of morphisms, functors, and natural compositions is associative
- * functors "distribute over composition", that is for any morphisms f and g in F's source category: F(g ∘ f) = F(g) ∘ F(f)
- * if η is a natural transformation from F to G, then for every f:C1→C2 in F and G's source category <b>C</b>: η[C2] ∘ F(f) = G(f) ∘ η[C1].
+* composition of morphisms, functors, and natural compositions is associative
+
+* functors "distribute over composition", that is for any morphisms `f` and `g` in `F`'s source category: <code>F(g ∘ f) = F(g) ∘ F(f)</code>
+
+* if <code>η</code> is a natural transformation from `G` to `H`, then for every <code>f:C1→C2</code> in `G` and `H`'s source category <b>C</b>: <code>η[C2] ∘ G(f) = H(f) ∘ η[C1]</code>.
+
+* <code>(η F)[E] = η[F(E)]</code>
+
+* <code>(K η)[E} = K(η[E])</code>
+
+* <code>((φ -v- η) F) = ((φ F) -v- (η F))</code>
Let's use the definitions of naturalness, and of composition of natural transformations, to establish two lemmas.
-Recall that join is a natural transformation from the (composite) functor MM to M. So for elements C1 in <b>C</b>, join[C1] will be a morphism from MM(C1) to M(C1). And for any morphism f:a→b in <b>C</b>:
+Recall that `join` is a natural transformation from the (composite) functor `MM` to `M`. So for elements `C1` in <b>C</b>, `join[C1]` will be a morphism from `MM(C1)` to `M(C1)`. And for any morphism <code>f:C1→C2</code> in <b>C</b>:
+
+<pre>
+ (1) join[C2] ∘ MM(f) = M(f) ∘ join[C1]
+</pre>
+
+Next, let <code>γ</code> be a transformation from `G` to `MG'`, and
+ consider the composite transformation <code>((join MG') -v- (MM γ))</code>.
- (1) join[b] ∘ MM(f) = M(f) ∘ join[a]
+* <code>γ</code> assigns elements `C1` in <b>C</b> a morphism <code>γ\*: G(C1) → MG'(C1)</code>. <code>(MM γ)</code> is a transformation that instead assigns `C1` the morphism <code>MM(γ\*)</code>.
-Next, consider the composite transformation ((join MQ') -v- (MM q)).
- q is a transformation from Q to MQ', and assigns elements C1 in <b>C</b> a morphism q*: Q(C1) → MQ'(C1). (MM q) is a transformation that instead assigns C1 the morphism MM(q*).
- (join MQ') is a transformation from MMMQ' to MMQ' that assigns C1 the morphism join[MQ'(C1)].
- Composing them:
- (2) ((join MQ') -v- (MM q)) assigns to C1 the morphism join[MQ'(C1)] ∘ MM(q*).
+* `(join MG')` is a transformation from `MM(MG')` to `M(MG')` that assigns `C1` the morphism `join[MG'(C1)]`.
-Next, consider the composite transformation ((M q) -v- (join Q)).
- (3) This assigns to C1 the morphism M(q*) ∘ join[Q(C1)].
+Composing them:
+
+<pre>
+ (2) ((join MG') -v- (MM γ)) assigns to C1 the morphism join[MG'(C1)] ∘ MM(γ*).
+</pre>
+
+Next:
+
+<pre>
+ (3) Consider the composite transformation <code>((M γ) -v- (join G))</code>. This assigns to C1 the morphism M(γ*) ∘ join[G(C1)].
+</pre>
+
+So for every element `C1` of <b>C</b>:
+
+<pre>
+ ((join MG') -v- (MM γ))[C1], by (2) is:
+ join[MG'(C1)] ∘ MM(γ*), which by (1), with f=γ*:G(C1)→MG'(C1) is:
+ M(γ*) ∘ join[G(C1)], which by 3 is:
+ ((M γ) -v- (join G))[C1]
+</pre>
+
+So our **(lemma 1)** is:
+
+<pre>
+ ((join MG') -v- (MM γ)) = ((M γ) -v- (join G)),
+ where as we said γ is a natural transformation from G to MG'.
+</pre>
-So for every element C1 of <b>C</b>:
- ((join MQ') -v- (MM q))[C1], by (2) is:
- join[MQ'(C1)] ∘ MM(q*), which by (1), with f=q*: Q(C1)→MQ'(C1) is:
- M(q*) ∘ join[Q(C1)], which by 3 is:
- ((M q) -v- (join Q))[C1]
-So our (lemma 1) is: ((join MQ') -v- (MM q)) = ((M q) -v- (join Q)), where q is a transformation from Q to MQ'.
+Next recall that `unit` is a natural transformation from `1C` to `M`. So for elements `C1` in <b>C</b>, `unit[C1]` will be a morphism from `C1` to `M(C1)`. And for any morphism <code>f:C1→C2</code> in <b>C</b>:
+
+<pre>
+ (4) unit[C2] ∘ f = M(f) ∘ unit[C1]
+</pre>
+Next:
-Next recall that unit is a natural transformation from 1C to M. So for elements C1 in <b>C</b>, unit[C1] will be a morphism from C1 to M(C1). And for any morphism f:a→b in <b>C</b>:
- (4) unit[b] ∘ f = M(f) ∘ unit[a]
+<pre>
+ (5) Consider the composite transformation ((M γ) -v- (unit G)). This assigns to C1 the morphism M(γ*) ∘ unit[G(C1)].
+</pre>
-Next consider the composite transformation ((M q) -v- (unit Q)). (5) This assigns to C1 the morphism M(q*) ∘ unit[Q(C1)].
+Next:
-Next consider the composite transformation ((unit MQ') -v- q). (6) This assigns to C1 the morphism unit[MQ'(C1)] ∘ q*.
+<pre>
+ (6) Consider the composite transformation ((unit MG') -v- γ). This assigns to C1 the morphism unit[MG'(C1)] ∘ γ*.
+</pre>
So for every element C1 of <b>C</b>:
- ((M q) -v- (unit Q))[C1], by (5) =
- M(q*) ∘ unit[Q(C1)], which by (4), with f=q*: Q(C1)→MQ'(C1) is:
- unit[MQ'(C1)] ∘ q*, which by (6) =
- ((unit MQ') -v- q)[C1]
-So our lemma (2) is: (((M q) -v- (unit Q)) = ((unit MQ') -v- q)), where q is a transformation from Q to MQ'.
+<pre>
+ ((M γ) -v- (unit G))[C1], by (5) =
+ M(γ*) ∘ unit[G(C1)], which by (4), with f=γ*:G(C1)→MG'(C1) is:
+ unit[MG'(C1)] ∘ γ*, which by (6) =
+ ((unit MG') -v- γ)[C1]
+</pre>
+
+So our **(lemma 2)** is:
+
+<pre>
+ (((M γ) -v- (unit G)) = ((unit MG') -v- γ)),
+ where as we said γ is a natural transformation from G to MG'.
+</pre>
-Finally, we substitute ((join Q') -v- (M q) -v- p) for q <=< p in the monad laws. For simplicity, I'll omit the "-v-".
+Finally, we substitute <code>((join G') -v- (M γ) -v- φ)</code> for <code>γ <=< φ</code> in the monad laws. For simplicity, I'll omit the "-v-".
- for all p,q,r in T, where p is a transformation from P to MP', q is a transformation from Q to MQ', R is a transformation from R to MR', and P'=Q and Q'=R:
+<pre>
+ for all φ,γ,ρ in T, where φ is a transformation from F to MF', γ is a transformation from G to MG', R is a transformation from R to MR', and F'=G and G'=R:
- (i) q <=< p etc are also in T
+ (i) γ <=< φ etc are also in T
==>
- (i') ((join Q') (M q) p) etc are also in T
+ (i') ((join G') (M γ) φ) etc are also in T
- (ii) (r <=< q) <=< p = r <=< (q <=< p)
+ (ii) (ρ <=< γ) <=< φ = ρ <=< (γ <=< φ)
==>
- (r <=< q) is a transformation from Q to MR', so:
- (r <=< q) <=< p becomes: (join R') (M (r <=< q)) p
- which is: (join R') (M ((join R') (M r) q)) p
+ (ρ <=< γ) is a transformation from G to MR', so:
+ (ρ <=< γ) <=< φ becomes: (join R') (M (ρ <=< γ)) φ
+ which is: (join R') (M ((join R') (M ρ) γ)) φ
substituting in (ii), and helping ourselves to associativity on the rhs, we get:
- ((join R') (M ((join R') (M r) q)) p) = ((join R') (M r) (join Q') (M q) p)
+ ((join R') (M ((join R') (M ρ) γ)) φ) = ((join R') (M ρ) (join G') (M γ) φ)
---------------------
which by the distributivity of functors over composition, and helping ourselves to associativity on the lhs, yields:
------------------------
- ((join R') (M join R') (MM r) (M q) p) = ((join R') (M r) (join Q') (M q) p)
+ ((join R') (M join R') (MM ρ) (M γ) φ) = ((join R') (M ρ) (join G') (M γ) φ)
---------------
- which by lemma 1, with r a transformation from Q' to MR', yields:
+ which by lemma 1, with ρ a transformation from G' to MR', yields:
-----------------
- ((join R') (M join R') (MM r) (M q) p) = ((join R') (join MR') (MM r) (M q) p)
+ ((join R') (M join R') (MM ρ) (M γ) φ) = ((join R') (join MR') (MM ρ) (M γ) φ)
- which will be true for all r,q,p just in case:
+ which will be true for all ρ,γ,φ just in case:
((join R') (M join R')) = ((join R') (join MR')), for any R'.
(ii') (join (M join)) = (join (join M))
- (iii.1) (unit P') <=< p = p
+ (iii.1) (unit F') <=< φ = φ
==>
- (unit P') is a transformation from P' to MP', so:
- (unit P') <=< p becomes: (join P') (M unit P') p
- which is: (join P') (M unit P') p
+ (unit F') is a transformation from F' to MF', so:
+ (unit F') <=< φ becomes: (join F') (M unit F') φ
+ which is: (join F') (M unit F') φ
substituting in (iii.1), we get:
- ((join P') (M unit P') p) = p
+ ((join F') (M unit F') φ) = φ
- which will be true for all p just in case:
+ which will be true for all φ just in case:
- ((join P') (M unit P')) = the identity transformation, for any P'
+ ((join F') (M unit F')) = the identity transformation, for any F'
which will in turn be true just in case:
(iii.1') (join (M unit) = the identity transformation
- (iii.2) p = p <=< (unit P)
+ (iii.2) φ = φ <=< (unit F)
==>
- p is a transformation from P to MP', so:
- unit <=< p becomes: (join P') (M p) unit
+ φ is a transformation from F to MF', so:
+ unit <=< φ becomes: (join F') (M φ) unit
substituting in (iii.2), we get:
- p = ((join P') (M p) (unit P))
+ φ = ((join F') (M φ) (unit F))
--------------
which by lemma (2), yields:
------------
- p = ((join P') ((unit MP') p)
+ φ = ((join F') ((unit MF') φ)
- which will be true for all p just in case:
+ which will be true for all φ just in case:
- ((join P') (unit MP')) = the identity transformation, for any P'
+ ((join F') (unit MF')) = the identity transformation, for any F'
which will in turn be true just in case:
(iii.2') (join (unit M)) = the identity transformation
+</pre>
Collecting the results, our monad laws turn out in this format to be:
- when p a transformation from P to MP', q a transformation from P' to MQ', r a transformation from Q' to MR' all in T:
+</pre>
+ when φ a transformation from F to MF', γ a transformation from F' to MG', ρ a transformation from G' to MR' all in T:
- (i') ((join Q') (M q) p) etc also in T
+ (i') ((join G') (M γ) φ) etc also in T
(ii') (join (M join)) = (join (join M))
(iii.1') (join (M unit)) = the identity transformation
(iii.2')(join (unit M)) = the identity transformation
+</pre>
In functional programming, instead of working with natural transformations we work with "monadic values" and polymorphic functions "into the monad" in question.
-For an example of the latter, let p be a function that takes arguments of some (schematic, polymorphic) type C1 and yields results of some (schematic, polymorphic) type M(C2). An example with M being the list monad, and C2 being the tuple type schema int * C1:
+For an example of the latter, let φ be a function that takes arguments of some (schematic, polymorphic) type C1 and yields results of some (schematic, polymorphic) type M(C2). An example with M being the list monad, and C2 being the tuple type schema int * C1:
- let p = fun c → [(1,c), (2,c)]
+ let φ = fun c → [(1,c), (2,c)]
-p is polymorphic: when you apply it to the int 0 you get a result of type "list of int * int": [(1,0), (2,0)]. When you apply it to the char 'e' you get a result of type "list of int * char": [(1,'e'), (2,'e')].
+φ is polymorphic: when you apply it to the int 0 you get a result of type "list of int * int": [(1,0), (2,0)]. When you apply it to the char 'e' you get a result of type "list of int * char": [(1,'e'), (2,'e')].
-However, to keep things simple, we'll work instead with functions whose type is settled. So instead of the polymorphic p, we'll work with (p : C1 → M(int * C1)). This only accepts arguments of type C1. For generality, I'll talk of functions with the type (p : C1 → M(C1')), where we assume that C1' is a function of C1.
+However, to keep things simple, we'll work instead with functions whose type is settled. So instead of the polymorphic φ, we'll work with (φ : C1 → M(int * C1)). This only accepts arguments of type C1. For generality, I'll talk of functions with the type (φ : C1 → M(C1')), where we assume that C1' is a function of C1.
-A "monadic value" is any member of a type M(C1), for any type C1. For example, a list is a monadic value for the list monad. We can think of these monadic values as the result of applying some function (p : C1 → M(C1')) to an argument of type C1.
+A "monadic value" is any member of a type M(C1), for any type C1. For example, a list is a monadic value for the list monad. We can think of these monadic values as the result of applying some function (φ : C1 → M(C1')) to an argument of type C1.