Then g = hf × f i
∗
∆.
Proof. From the above hf × f i
∗
∆ ⊑ g ◦ g
−1
⊑ g.
[FIXME: Funcoids and reloids are confused.]
It’s remainded to prove g ⊑ hf × f i
∗
∆.
[FIXME: Possible er rors.]
Suppose there is U ∈ xyGR hf × f i
∗
∆ such that U
GR g.
Then {V \ U | V ∈ GR g} = g \ U would be a proper filter.
Thus by reflexivity hf × f i
∗
(g \ U)
0.
By compactness of f × f, Cor hf × f i
∗
(g \ U )
0.
Suppose ↑{(x; x)} ⊑ hf × f i
∗
(g \ U); then g \ U
hf
−1
× f
−1
i{(x; x)}; U ⊏ hf
−1
× f
−1
i{(x;
x)} ⊑ hf
−1
× f
−1
i∆ what is impossible.
Thus there exist x
y such that {(x; y)} ⊑ Cor hf × f i
∗
(g \ U). Thus {(x; y)} ⊑ hf × f i
∗
g.
Thus by the lemma {(x; y)} ⊑ ∆ w hat is impossible. So U ∈ GR g.
We have xyGR hf × f i
∗
∆ ⊆ GR g; hf × f i
∗
∆ ⊒ g.
Corollary 18. Let f is a T
1
-separable (the same as T
2
for symmetric transitive) compact funcoid
and g is a uniform space (reflexive, symm etric, and transitive endoreloid) such that (FCD)g = f.
Then g = hf × f i
∗
∆.
An (incomplete) attempt to prove one more theorem follows:
Theorem 19. Le t µ and ν be uniform spaces, (FCD)µ be a compact funcoid. Then a map f i s a
continuous map from (FCD)µ to (FCD)ν iff f is a (uniformly) continuou s map from µ to ν.
Proof. [FIXME: errors in this proof.]
We have µ = h(FCD)µ × (FCD)µi↑
RLD
∆
f ∈ C
?
((FCD) µ; (FCD)ν). Then
f × f ∈ C
?
((FCD)(µ × µ); (FCD)(ν × ν))
(f × f) ◦ ( FCD)(µ × µ) ⊑ (FCD)(ν × ν) ◦ (f × f )
For eve ry V ∈ GR(ν × ν) we have hg
−1
iV ∈ h(FCD)(µ × µ)i{y} for some y.
hg
−1
iV ∈ h(FCD)µ × (FCD)µi↑
RLD
∆ = GR µ
hgihg
−1
iV ⊑ V
We need to prove f ∈ C(µ; ν) that is ∀p ∈ GR ν ∃q ∈ GR µ: hf iq ⊑ p. But this follows from the
above.
Bib liography
[1] Victor Porton. Categorical product of funcoids. At http://www.mathematics21.org/binaries/product.pdf.
[2] Victor Porton. Algebraic General Topology. Volume 1. 2013.
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