8. CARTESIAN CLOSEDNESS 16
8.2. Conjectures.
Conjecture 2017. The categories Fcd and Rld are cartesian closed (actually
two conjectures).
http://mathoverﬂow.net/questions/141615/how-to-prove-that-there-are-no-exponential-object-in-a-category
suggests to investigate colimits to prove that there are no exponential object.
Our purpose is to prove (or disprove) that categories Dig, Fcd, and Rld are
cartesian closed. Note that they have ﬁnite (and even inﬁnite) products is already
proved.
Alternative way to prove: you can prove that the functor − × B is left adjoint
to the exponentiation −
B
where the counit is given by the evaluation map.
See http://www.springer.com/us/book/9780387977102 for another way to
prove Cartesian closedness.
8.3. Category Dig is cartesian closed. Category of digraphs is the sim-
plest of our three categories and it is easy to demonstrate that it is cartesian closed.
I demonstrate cartesian closedness of Dig mainly with the purpose to show a pat-
tern similarly to which we may probably demonstrate our two other categories are
cartesian closed.
Let G and H be graphs:
• Ob MOR(G, H) = (Ob H)
Ob G
;
• (f, g) ∈ GR MOR(G, H) ⇔ ∀(v, w) ∈ GR G : (f(v), g(w)) ∈ GR H for
every f, g ∈ Ob MOR(G, H) = (Ob H)
Ob G
;
GR 1
MOR(B,C)
= id
Ob MOR(B,C)
= id
(Ob H)
Ob G
Equivalently
(f, g) ∈ GR MOR(G, H) ⇔ ∀(v, w) ∈ GR G : g ◦ {(v, w)} ◦ f
−1
⊆ GR H
(f, g) ∈ GR MOR(G, H) ⇔ g ◦ (GR G) ◦ f
−1
⊆ GR H
(f, g) ∈ GR MOR(G, H) ⇔ hf ×
(C)
gi GR G ⊆ GR H
The transposition (the isomorphism) is uncurrying.
∼ f = λa ∈ Zλy ∈ A : f(a, y) that is (∼ f)(a)(y) = f(a, y).
(−f)(a, y) = f(a)(y)
If f : A × B → C then ∼ f : A → MOR(B, C)
”Proposition” Transposition and its inverse are morphisms of Dig.
”Proof” It follows from the equivalence ∼ f : A → MOR(B, C) ⇔ ∀x, y :
(xAy ⇒ (∼ f)x(MOR(B, C))(∼ f )y) ⇔
∀x, y : (xAy ⇒ ∀(v, w) ∈ B : ((∼ f)xv, (∼ f)yw) ∈ C) ⇔
∀x, y, v, w : (xAy ∧ vBw ⇒ ((∼ f)xv, (∼ f )yw) ∈ C) ⇔
∀x, y, v, w : ((x, v)(A × B)(y, w) ⇒ (f (x, v), f(y, w)) ∈ C) ⇔ f : A × B → C.
Evaluation ε : MOR(G, H) × G → H is deﬁned by the formula:
Then evaluation is ε
B,C
= −(1
MOR(B,C)
).
So ε
B,C
(p, q) = (−(1
MOR(B,C)
))(p, q) = (1
MOR(B,C)
)(p)(q) = p(q).
”Proposition” Evaluation is a morphism of Dig.
”Proof” Because ε
B,C
(p, q) = −(1
MOR(B,C)
).
It remains to prove: * ε
B,C
◦ (∼ f × 1
A
) = f for f : A → B × C; * ∼
(ε
B,C
◦ (g × 1
A
)) = g for g : A → MOR(B, C).
”Proof” ε
B,C
(∼ f × 1
A
)(a, p) = ε
B,C
((∼ f )a, p) = (∼ f )ap = f(a, p). So
ε
B,C
◦ (∼ f × 1
A
) = f.
∼ (ε
B,C
◦ (g × 1
A
))(p)(q) = (ε
B,C
◦ (g × 1
A
))(p, q) = ε
B,C
(g × 1
A
)(p, q) =
ε
B,C
(gp, q) = g(p)(q). So ∼ (ε
B,C
◦ (g × 1
A
)) = g.
8.4. By analogy with the proof that Dig is cartesian closed. The most
obvious way for proof attempt that Fcd is cartesian closed is an analogy with the
proof that Dig is cartesian closed.