To have a category, the elements and morphisms have to satisfy some constraints:
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- (i) the class of morphisms has to be closed under composition: where f:C1→C2 and g:C2→C3, g º f is also a morphism of the category, which maps C1→C3.
+ (i) the class of morphisms has to be closed under composition: where f:C1→C2 and g:C2→C3, g ∘ f is also a morphism of the category, which maps C1→C3.
(ii) composition of morphisms has to be associative
- (iii) every element E of the category has to have an identity morphism 1<sub>E</sub>, which is such that for every morphism f:C1→C2: 1<sub>C2</sub> o f = f = f o 1<sub>C1</sub>
+ (iii) every element E of the category has to have an identity morphism 1<sub>E</sub>, which is such that for every morphism f:C1→C2: 1<sub>C2</sub> ∘ f = f = f ∘ 1<sub>C1</sub>
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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.
(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 o f) = F(g) o F(f)
+ (iv) "distribute over composition", that is for any morphisms f and g in <b>C</b>: F(g ∘ f) = F(g) ∘ F(f)
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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`.
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:
- for every morphism f:C1→C2 in <b>C</b>: η[C2] o G(f) = H(f) o η[C1]
+ for every morphism f:C1→C2 in <b>C</b>: η[C2] ∘ G(f) = H(f) ∘ η[C1]
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)`.
`(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`:
- φ[C2] o H(f) o η[C1] = φ[C2] o H(f) o η[C1]
+ φ[C2] ∘ H(f) ∘ η[C1] = φ[C2] ∘ H(f) ∘ η[C1]
by naturalness of φ, is:
- φ[C2] o H(f) o η[C1] = J(f) o φ[C1] o η[C1]
+ φ[C2] ∘ H(f) ∘ η[C1] = J(f) ∘ φ[C1] ∘ η[C1]
by naturalness of η, is:
- φ[C2] o η[C2] o G(f) = J(f) o φ[C1] o η[C1]
+ φ[C2] ∘ η[C2] ∘ G(f) = J(f) ∘ φ[C1] ∘ η[C1]
-Hence, we can define `(φ -v- η)[x]` as: φ[x] o η[x]` and rely on it to satisfy the constraints for a natural transformation from `G` to `J`:
+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`:
- (φ -v- η)[C2] o G(f) = J(f) o (φ -v- η)[C1]
+ (φ -v- η)[C2] ∘ G(f) = J(f) ∘ (φ -v- η)[C1]
An observation we'll rely on later: given the definitions of vertical composition and of how natural transformations compose with functors, it follows that:
`(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:
- (φ -h- η)[C1] = L(η[C1]) o ψ[G(C1)]
- = ψ[H(C1)] o K(η[C1])
+ (φ -h- η)[C1] = L(η[C1]) ∘ ψ[G(C1)]
+ = ψ[H(C1)] ∘ K(η[C1])
Horizontal composition is also associative, and has the same identity as vertical composition.
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> o f = f = f o 1<sub>C1</sub>.
+ 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 o f) = F(g) o 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] o F(f) = G(f) o η[C1].
+ * 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].
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>:
- (1) join[b] o MM(f) = M(f) o join[a]
+ (1) join[b] ∘ MM(f) = M(f) ∘ join[a]
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)] o MM(q*).
+ (2) ((join MQ') -v- (MM q)) assigns to C1 the morphism join[MQ'(C1)] ∘ MM(q*).
Next, consider the composite transformation ((M q) -v- (join Q)).
- (3) This assigns to C1 the morphism M(q*) o join[Q(C1)].
+ (3) This assigns to C1 the morphism M(q*) ∘ join[Q(C1)].
So for every element C1 of <b>C</b>:
((join MQ') -v- (MM q))[C1], by (2) is:
- join[MQ'(C1)] o MM(q*), which by (1), with f=q*: Q(C1)→MQ'(C1) is:
- M(q*) o join[Q(C1)], which by 3 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 f:a→b in <b>C</b>:
- (4) unit[b] o f = M(f) o unit[a]
+ (4) unit[b] ∘ f = M(f) ∘ unit[a]
-Next consider the composite transformation ((M q) -v- (unit Q)). (5) This assigns to C1 the morphism M(q*) o unit[Q(C1)].
+Next consider the composite transformation ((M q) -v- (unit Q)). (5) This assigns to C1 the morphism M(q*) ∘ unit[Q(C1)].
-Next consider the composite transformation ((unit MQ') -v- q). (6) This assigns to C1 the morphism unit[MQ'(C1)] o q*.
+Next consider the composite transformation ((unit MQ') -v- q). (6) This assigns to C1 the morphism unit[MQ'(C1)] ∘ q*.
So for every element C1 of <b>C</b>:
((M q) -v- (unit Q))[C1], by (5) =
- M(q*) o unit[Q(C1)], which by (4), with f=q*: Q(C1)→MQ'(C1) is:
- unit[MQ'(C1)] o q*, which by (6) =
+ 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'.
A natural transformation t assigns to each type C1 in <b>C</b> a morphism t[C1]: C1→M(C1) such that, for every f:C1→C2:
- t[C2] o f = M(f) o t[C1]
+ t[C2] ∘ f = M(f) ∘ t[C1]
The composite morphisms said here to be identical are morphisms from the type C1 to the type M(C2).