A CHARACTERIZATION OF CLOSED MAPS USING THE WHYBURN CONSTRUCTION

In this paper we modify the Whyburn construction for a continuous function f:X→Y. If the range is first countable, we get a characterization of closed maps; namely, the constructions are the same if and only if the map is closed.


INTRODUCTION.
Let f X Y be continuous and let X and Y be Hausdorff.In [3] Whyburn defined the unified space Z to be the disjoint union of X and Y with a set open in Z if and only if Q  X is open in X, Q Y is open in Y, and for any compact Kc ( N Y, f-l(K) (  COROLLARY.Any continuous function from a Hausdorff space into the reals {or any metric space) is closed if and only if the Whyburn space and the modified Whyburn space are the same.Z. PRELIMINARIES.
Arguments similar to those of Whyburn's show that W is a T topological space containing X as an open subspace and Y as a closed subspace.However, Y. 0. STALLINGS just as in the Whyburn space, W need not be Hausdorff.Whyburn showed that Z is Hausdorff if X is locally compact.Asking when W is I-Iausdorff led to the following definitions and propositions: DEFINITION 2. 1.Let f X--Y be continuous.Then A c-X is fiber compact if and only if A is closed and for all y e f(A), f-l(y) A is compact.
Also, X is locally fiber compact if every point has a neighborhood whose closure is fiber compact.
PROOF.The only interesting case is when p is in X and f(p) q.Since X is locally fiber compact, there exists a U open in X such that p is in U and U is fiber compact.Hence U is open in W and W U is a neighborhood of q by Proposition Z. 2. We define, as did Whyburn, a retraction r W--Y to be f on X and the identity on Y.The following results parallel those of Whyburn's for r Z Y.
The proof is omitted.
PROPOSITION 2.4.The map r %%r y is continuous, has compact fibers The next proposition shows that some of the properties mentioned above actually characterize the modified Whyburn construction.This proposition is similar to a theorem about the Whyburn construction proved by Dickrnan [I].
PROPOSITION 2.5.Let r S Y be a retraction with compact fibers from a Hausdorff space onto a regular subspace.Let X S Y and f r IX" If fiber compact subsets of X are closed in S, then the modified Whyburn space for f X Y is homeomorphic to S.
PROOF.Let V be the modified Vhyburn space for f X Y.
set in S and is contained in Q. Suppose x e Q fl Y. Then since Y is regular, we can find a neighborhood V of x such that x Vc VcQ fl Y. Let f I(V) Q B.
-1 Then B is fiber compact and so S B is open in S. Let U (S B) fl r (V).
Then U contains x, is open in S, and is contained in 3. MAIN THEOREM.
We now state and prove the major theorem of this paper which allows us to determine when W and Z are the same.PROOF.Assume f is closed, Q is open in W and K is any compact sub- Then A is closed and f(A) is a closed subset of K and hence is compact.Then f[A A f(A) is a continuous, closed sur- jection with compact fibers and therefore is a perfect map.By [2, Theorem 5.3] Now assume that Z and W are equal.Let A be a closed set in X.
Suppose that y is a limit point of f{A).Since Y is first countable and Hausdorff, there exists a sequence of distinct points {yn} c f(A) which converges to y.So we may choose a sequence {Xn} in A such that f(Xn) Yn" Let B {Xn}.Now suppose B has no limit points.Then B is closed in X.Since for any Yn e f(B), f-l(yn) N B {Xn} B is fiber compact and thus ) (Z B) B is compact.Since B is also infinite it must have a limit point, contradicting our assumption.Hence B has a limit point, say x.
Suppose that f(x) z % y.Then we can find disjoint neighborhoods V of z and U of y.Since {yn} converges to y, there exists an N such that for every n _> N, Yn e U.However, since x is a limit point of B, we have an integer -1 m > N such that Xm e f (V).Hence f(Xm) Ym is in both U and V which is impossible.Hence f(x) y.Since BoA and A is closed, x e A and there.. fore f(A) is closed.
Notice that Y being first countable is a necessary hypothesis for the preceding theorem.The following is an example to illustrate this.
Let X. [0, 1) for all 1,2,3 Then let X be the disjoint union of these X.'s.Let Y X U p where p is not in X. Define Vc Y to be open if and only if 1) V is an open set contained in X or 2) If p e V, then there exists a finite set of indices such that if e {i 1,..., in} then Xi V X.1 and if t {i 1, in} then X.1 V is compact.
The inclusion map from X to Y is not closed, Y is not first countable at p and yet W and Z are the same.

THEOREM 3 .
I. Let f X Y be continuous, X and Y be Hausdorff, and let Y be first countable.Then Z and W are equal if and only if f is a closed mapping.
is compact.In this paper we modify the topology on X U Y by defining to be open if and only if C) X is open in X, C) N Y is open in Y, and for any point p ( Y, f-l{p) Q is compact.We denote the modified Whyburn space by It is obvious that any set open in Z is open in W. We will show that if f is closed, the topologies are in fact the same and if Y is first countable, then Z and W being the same implies that f is closed.This will yield the following corollary: