Exercise: If
f = m ∙ e
is an idempotent satisfying
f ∙ f = f
and
e ∙ m = 1P
(where ‘∙’ denotes composition), then the section
m: P –> A
is an equalizer of the parallel pair of maps
f, 1A: A –> A
(Conceptual Mathematics, p. 293).
Definition: An equalizer of a parallel pair of maps
f, g: A –> B
is a map
i: X –> A
satisfying
f ∙ i = g ∙ i
and every map
j: Y –> A
satisfying
f ∙ j = g ∙ j
is uniquely included in
i: X –> A
i.e. there is exactly one map
k: Y –> X
such that
j = i ∙ k
To show that
m: P –> A
is an equalizer of
f, 1A: A –> A,
given
f = m ∙ e
f ∙ f = f
e ∙ m = 1P,
we have to first show that
f ∙ m = 1A ∙ m
i.e.
f ∙ m = m
f ∙ m = (m ∙ e) ∙ m = m ∙ (e ∙ m) = m ∙ 1P = m
Next, we have to show that if any map
j: Y –> A
satisfies
f ∙ j = j
then there is exactly one map
k: Y –> P
satisfying
j = m ∙ k
First, we need a map from Y to P. Given
j: Y –> A
and
e: A –> P
we can take the composite map
e ∙ j: Y –> A –> P
as the map k from Y to P i.e.
k = e ∙ j
and see if
m ∙ k = j
i.e. if
m ∙ (e ∙ j) = j
m ∙ (e ∙ j) = (m ∙ e) ∙ j = f ∙ j = j
(since we are given f ∙ j = j).
Finally, we have to show that
k: Y –> P
satisfying
m ∙ k = j
is unique i.e. if there is another map
k’: Y –> P
satisfying
m ∙ k’ = j
then
k’ = k
Given
m ∙ k’ = j = m ∙ k
post-composing with
e: A –> P
on both sides of
m ∙ k’ = m ∙ k
we find
e ∙ (m ∙ k’) = e ∙ (m ∙ k)
(e ∙ m) ∙ k’ = (e ∙ m) ∙ k
1P ∙ k’ = 1P ∙ k
k’ = k
Thus the section
m: P –> A
of the splitting of an idempotent
f = m ∙ e: A –> P –> A
is an equalizer of
f, 1A: A –> A
Tailpiece: An equalizer of two idempotents
f, f’: A –> A
with a common section i.e.
f = m ∙ e
f’ = m ∙ e’
is the common section
m: P –> A
True or false? (For some reason, I’m not my usual verbose self :( I’ll back soon to spice-up the story all the way to cohesion vs. quality of reflexive graphs vs. idempotents via points and pieces functors constructed in terms of equalizers and coequalizers.)
Science of Knowing 