By Allan J. Sieradski

This article is an creation to topology and homotopy. issues are built-in right into a coherent complete and built slowly so scholars are usually not crushed. the 1st half the textual content treats the topology of whole metric areas, together with their hyperspaces of sequentially compact subspaces. the second one half the textual content develops the homotopy type. there are lots of examples and over 900 workouts, representing quite a lot of trouble. This publication may be of curiosity to undergraduates and researchers in arithmetic.

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Thus, a = ν(a) ∈ S. 1. Corollary. Each frame congruence E can be expressed as E= {∇a ∩ Δb | aEb}. 7. Open and closed localic maps We will analyze the counterparts of open continuous maps and closed continuous maps in classical spaces. 1. Lemma. Let f : L → M be a localic map and let S be a sublocale. Then we have for the congruences associated with S and f [S] the formula aEf [S] b iﬀ f ∗ (a)ES f ∗ (b). 7. Open and closed localic maps 37 Proof. 3. We have aEf [S] b iﬀ ∀s ∈ S, (a ≤ f (s) iﬀ b ≤ f (s)) iﬀ ∀s ∈ S, (f ∗ (a) ≤ s iﬀ f ∗ (b) ≤ s) iﬀ f ∗ (a)ES f ∗ (b).

The join is even more entangled. Compare this with the lattice operations in S (L) ! The translation of sublocale homomorphisms to sublocales as above, and vice versa, is as follows: h → h∗ [M ] for an onto h : L → M and h∗ its right adjoint, and S → jS∗ : L → S for jS : S ⊆ L. 2. Frame congruences. Using frame congruences (that is, equivalences respecting all joins and ﬁnite meets) is more handy. A sublocale homomorphism g : L → M 5. Alternative representations of sublocales 31 induces a frame congruence Eg = {(x, y) | g(x) = g(y)} and a frame congruence gives rise to the sublocale homomorphism (x → Ex) : L → L/E (where L/E denotes the quotient frame deﬁned by the congruence E, just as quotients are always deﬁned for algebraic systems, and Ex denotes the E-class {y | (y, x) ∈ E} of x ∈ L).

Lemma. (a) For every a, b ∈ L we have ν(a → ν(b)) = a → ν(b). (b) For every a, b ∈ L, b → ν(a) = ν(b) → ν(a). Proof. 1 and (N2), (N3), we obtain ν(a → ν(b)) ∧ a ≤ ν(a → ν(b)) ∧ ν(a) = ν((a → ν(b)) ∧ a) ≤ ν(ν(b)) = ν(b). Thus, by the basic Heyting formula, ν(a → ν(b)) ≤ a → ν(b). Finally, use (N1) again. 1, b → ν(a) ∧ b ≤ ν(a) and, by(N4) and (N3), ν(b → ν(a)) ∧ ν(b) ≤ ν(a). By (H) and (N1), b → ν(a) ≤ ν(b) → ν(a). Since u → v is antitone in the ﬁrst variable, we also have ν(b) → ν(a) ≤ b → ν(a).