已知X,Y为正规空间，求证粘合空间X∪_f Y的正规性

Alright, let's work through proving that the adjunction space (X \cup_f Y) is normal when (X) and (Y) are normal spaces. First, let's align on the problem statement clearly (both cases you mentioned are essentially the same—we can always assume (X) and (Y) are disjoint by replacing (Y) with a homeomorphic copy if needed):

Given normal spaces (X) and (Y), a closed subset (A \subseteq X), and a continuous function (f: A \to Y), the adjunction space (Z = X \cup_f Y) is the quotient space of (X \cup Y) under the equivalence relation (x \sim f(x)) for all (x \in A) (all other points are only equivalent to themselves). We need to show (Z) is a normal space (disjoint closed subsets can be separated by disjoint open subsets).

Let (q: X \cup Y \to Z) denote the quotient map. Recall that a subset (S \subseteq Z) is closed if and only if (q^{-1}(S)) is closed in (X \cup Y) (a core property of quotient maps). Also, since (A) is closed in (X), (q) is a closed quotient map—the image of any closed set in (X \cup Y) is closed in (Z). This will be useful later.

Step 1: Lift Closed Subsets to (X \cup Y) #

Take two disjoint closed subsets (C, D \subseteq Z). Define:

• (C_X = q^{-1}(C) \cap X), (C_Y = q^{-1}(C) \cap Y)
• (D_X = q^{-1}(D) \cap X), (D_Y = q^{-1}(D) \cap Y)

By definition of quotient closed sets:

• (C_X, D_X) are closed in (X); (C_Y, D_Y) are closed in (Y)
• (C_X \cap D_X = \emptyset), (C_Y \cap D_Y = \emptyset)
• (C_X \cap f^{-1}(D_Y) = \emptyset) and (D_X \cap f^{-1}(C_Y) = \emptyset) (if (x) were in both, (q(x) = q(f(x))) would be in (C \cap D), which is impossible)

Step 2: Separate Lifted Subsets in (X) #

Consider the closed subsets (C_X \cup f^{-1}(C_Y)) and (D_X \cup f^{-1}(D_Y)) in (X). These are disjoint (verify using the bullet points above—all pairwise intersections are empty). Since (X) is normal, there exist disjoint open sets (U_X, V_X \subseteq X) such that:

• (C_X \cup f^{-1}(C_Y) \subseteq U_X)
• (D_X \cup f^{-1}(D_Y) \subseteq V_X)

Step 3: Separate Corresponding Subsets in (Y) #

Now look at (Y). Define two closed subsets:

• (E_Y = C_Y \cup f(X \setminus U_X))
• (F_Y = D_Y \cup f(X \setminus V_X))

These are closed because:

• (X \setminus U_X) is closed in (X), so (X \setminus U_X \cap A) is closed in (A); since (f) is continuous, (f(X \setminus U_X \cap A) = f(X \setminus U_X)) is closed in (Y)
• (C_Y) is already closed in (Y), so their union is closed

Also, (E_Y) and (F_Y) are disjoint:

• If (y \in E_Y \cap F_Y), either (y \in C_Y \cap D_Y) (impossible), or (y = f(x_1) = f(x_2)) where (x_1 \in X \setminus U_X) and (x_2 \in X \setminus V_X). But (x_1 \in X \setminus U_X) implies (x_1 \notin f^{-1}(C_Y)), so (f(x_1) \notin C_Y); similarly (f(x_2) \notin D_Y). Moreover, (x_1 \in X \setminus U_X) means (x_1 \notin U_X), so (x_1 \in V_X) (since (U_X) and (V_X) are disjoint), which contradicts (x_1 \in X \setminus V_X).

Since (Y) is normal, we can separate (E_Y) and (F_Y) with disjoint open sets (U_Y, V_Y \subseteq Y) where:

• (E_Y \subseteq U_Y)
• (F_Y \subseteq V_Y)

Step 4: Define Open Sets in (Z) #

Now define:

• (U = q(U_X \cup U_Y))
• (V = q(V_X \cup V_Y))

We need to confirm three things:

1. (U) and (V) are open in (Z): Since (q) is a closed map, the complement of (U) is (q((X \setminus U_X) \cup (Y \setminus U_Y))), which is closed (as the image of a closed set). Hence (U) is open; same for (V).
2. (C \subseteq U) and (D \subseteq V): (q^{-1}(C) = C_X \cup C_Y), and (C_X \subseteq U_X), (C_Y \subseteq U_Y), so (q(C_X \cup C_Y) = C \subseteq U). The same logic applies to (D \subseteq V).
3. (U \cap V = \emptyset): Suppose (z \in U \cap V). Then (z = q(a) = q(b)) where (a \in U_X \cup U_Y) and (b \in V_X \cup V_Y).
   • If (a \in U_X) and (b \in V_X): (U_X) and (V_X) are disjoint, so (a = b) is impossible. The only other way (q(a) = q(b)) is if (a \in A) and (b = f(a)) (or vice versa), but (b = f(a) \in U_Y) and (b \in V_Y), which are disjoint—contradiction.
   • If (a \in U_Y) and (b \in V_Y): (U_Y) and (V_Y) are disjoint, so this is impossible.
   • If (a \in U_X) and (b \in V_Y): Then (b = f(a)) (since (q(a) = q(b))). But (a \in U_X \subseteq X \setminus V_X), so (f(a) \in f(X \setminus V_X) \subseteq F_Y \subseteq V_Y). However, (a \in U_X) implies (a \notin X \setminus U_X), so (f(a) \notin f(X \setminus U_X) \subseteq E_Y \subseteq U_Y). But (q(a) = f(a) \in U) means (f(a) \in U_Y), which contradicts (f(a) \in V_Y) (since (U_Y \cap V_Y = \emptyset)).

All cases lead to a contradiction, so (U \cap V = \emptyset).

Conclusion #

We've shown that any two disjoint closed subsets of (Z = X \cup_f Y) can be separated by disjoint open subsets. Hence (Z) is a normal space.

内容的提问来源于stack exchange，提问作者mbrg