§ Quantifiers as adjoints
 Consider
S(x, y) ⊂ X × Y
, as a relation that tells us when (x, y)
is true.  We can then interpret
∀x, S(x, y)
to be a subset of Y
, that has all the elements such that this predicate holds. ie, the set { y : Y  ∀ x, S(x, y) }
.  Similarly, we can interpret
∃x, S(x, y)
to be a subset of Y
given by { y : Y  ∃ x, S(x, y) }
.  We will show that these are adjoints to the projection
π: X × Y → Y
.  Treat
P(S)
to be the boolean algebra of all subsets of S
, and similarly P(Y)
.  Then we can view
P(S)
and P(Y)
to be categories, and we have the functor π: P(S) → P(Y)
.  Recall that in this boolean algebra and arrow
a → b
denotes a subset relation a ⊆ b
.
§ A first try: direct image, find right adjoint
 Suppose we want to analyze when
π T ⊆ Z
, with the hopes of getting some condition when T ⊆ ? Z
where ?
is some tobedefined adjoint to π
.  See that
π T ⊆ Z
then means ∀ (x, y) ∈ T, y ∈ Z
.
T
t t t
t t t

v
tttt π(T)
zzzzzzzzzZ
 Suppose we build the set
Q(Z) ≡ { (x, y) ∈ S : y ∈ Z }
. That is to say, Q ≡ π⁻¹(Z)
. ( Q
for inverse of P
).  Then, it's clear that we have
π T ⊂ Z
implies that T ⊆ Q(Z)
[almost by definition ].  However, see that this
Q(Z)
construction goes in the wrong direction; we want a functor from P(S)
to P(Y)
, which projects out a variable via ∃ / ∀
. We seem to have built a functor in the other direction, from P(Y)
to P(S)
.  Thus, what we must actually do is to reverse the arrow
π: S ⊆ X × Y → Y
, and rather we must analyze π⁻¹
itself, because its adjoints will have the right type.  However, now that we've gotten this far, let's also analyze left adjoints to
π
.
§ Direct image, left adjoint
 Suppose that
Z ⊆ π T
. This means that for every y ∈ Z
, there is some x_y
such that (x_y, y) ∈ T
T
t t t
t t t

v
tttt π(T)
zzZ
 I want to find an operation
?
such that ? Z ⊆ T
.  One intuitive operation that comes to mind to unproject, while still reminaing a subset, is to use
π⁻¹(Z) ∩ T
. This would by construction have that π⁻¹(Z) ∩ T ⊆ T
.  Is this an adjoint? we'll need to check the equation
:)
.
§ Inverse image, left adjoint.
 Suppose we consider
π⁻¹ = π* : P(Y) → P(S)
.  Now, imagine we have
π*(Z) ⊆ T
.
S


tttt
tztt
tztt T
tztt
^^
 π*(Z)
zzZ
 In this case, we can say that for each
z ∈ Z
, for all x ∈ X
such that (x, z) ∈ S
, we had (x, z) ∈ T
.  Consider the set
∀ T ≡ { y ∈ T: ∀ x, (x, y) ∈ S => (x, y) ∈ T}
.  Thus, we can say that
π*(Z) ⊂ T
iff Z ⊂ ∀ T
.  Intuitively,
T ⊂ π*(π(T))
, so it must be "hard" for the inverse image of a set Z
( π*(Z)
) to be contained in the set T
, because inverse images cannot shrink the size.  Furthermore, it is the right adjoint to
π*(Z)
because the ???