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Applied General Topology

c© Universidad Politécnica de Valencia

Volume 9, No. 2, 2008

pp. 263-280

On the continuity of factorizations

W. W. Comfort, Ivan S. Gotchev and Luis Recoder-Núñez∗

Abstract. Let {Xi : i ∈ I} be a set of sets, XJ :=
∏

i∈J
Xi when

∅ 6= J ⊆ I; Y be a subset of XI , Z be a set, and f : Y → Z. Then f
is said to depend on J if p, q ∈ Y , pJ = qJ ⇒ f (p) = f (q); in this case,
fJ : πJ [Y ] → Z is well-defined by the rule f = fJ ◦ πJ |Y .
When the Xi and Z are spaces and f : Y → Z is continuous with Y
dense in XI , several natural questions arise:

(a) does f depend on some small J ⊆ I?
(b) if it does, when is fJ continuous?

(c) if fJ is continuous, when does it extend to continuous fJ :
XJ → Z?

(d) if fJ so extends, when does f extend to continuous f : XI →
Z?

(e) if f depends on some J ⊆ I and f extends to continuous

f : XI → Z, when does f also depend on J?

The authors offer answers (some complete, some partial) to some of
these questions, together with relevant counterexamples.
Theorem 1. f has a continuous extension f : XI → Z that depends
on J if and only if fJ is continuous and has a continuous extension
fJ : XJ → Z.
Example 1. For ω ≤ κ ≤ c there are a dense subset Y of [0, 1]κ and
f ∈ C(Y, [0, 1]) such that f depends on every nonempty J ⊆ κ, there
is no J ∈ [κ]<ω such that fJ is continuous, and f extends continuously
over [0, 1]κ.
Example 2. There are a Tychonoff space XI , dense Y ⊆ XI , f ∈
C(Y ), and J ∈ [I]<ω such that f depends on J, πJ [Y ] is C-embedded
in XJ , and f does not extend continuously over XI .

2000 AMS Classification: Primary 54B10, 54C20; Secondary 54C45.

Keywords: Product space, dense subspace, continuous factorization, contin-
uous extensions of maps, C(X).

∗The authors gratefully thank Gary Gruenhage for helpful e-mail correspondence.



264 W. W. Comfort, I. S. Gotchev and L. Recoder-Núñez

1. Introduction

In addition to the notation given in the Abstract, we adopt these conven-
tions. ω is (the cardinality of) the set of non-negative integers and I is an index
set (usually infinite). α, κ, and λ are cardinals and [I]<κ := {J ⊆ I : |J| < κ}.
(XI )κ denotes XI :=

∏
i∈I Xi with the κ-box topology (so (XI )ω = XI ) and

Σλ(p) := {x ∈ XI : |{i ∈ I : xi 6= pi}| < λ} whenever p ∈ XI . By a (canonical)
basic open set in (XI )κ we mean a set of the form U = UI = Πi∈I Ui with Ui
open in Xi and with R(U) := {i ∈ I : Ui 6= Xi} ∈ [I]

<κ. (In the terminology
of [4], R(U) is the restriction set of the (basic) open set U.) The symbol R
denotes the real line with its usual topology, the cardinality of R is denoted by
c, the cardinality of the set X is denoted by |X|, and the closure of X by X.

The weight of a space X is w(X) := min{|B| : B is a base for X} + ω, the
density of X is d(X) := min{|D| : D is dense in X} + ω, χ(x,X) denotes the
character (i.e., the local weight) of the point x in the space X, and χ(X) :=
sup{χ(x,X) : x ∈ X}. Finally, a pairwise disjoint collection of nonempty open
sets in X is called a cellular family and the cellularity of X is c(X) := sup{|U| :
U is a cellular family in X} + ω.

For spaces Y and Z we denote by C(Y,Z) the set of continuous functions
from Y into Z. We write C(X) := C(X, R) and (in contrast with the convention
used in [8] and elsewhere) we write C∗(X) := C(X, [0, 1]). A subspace Y of
a space X is C(Z)-embedded if every f ∈ C(Y,Z) extends to f ∈ C(X,Z);
then as usual [8], a C(Z)-embedded space Y ⊆ X with Z = R is said to be
C-embedded; if Z = [0, 1] then Y is said to be C∗-embedded.

Our spaces are not subjected to any standing separation hypothesis. When
a specific property is wanted, as in Remark 3.4, Section 4, and Theorems 5.5
and 5.6, we state it explicitly.

Definition 1.1. When XI , Y , Z and f are as in the Abstract and f depends
on J ⊆ I, the function fJ : πJ [Y ] → Z (defined by the relation f = fJ ◦ πJ |Y )
is a factorization of f.

For additional topological definitions not given above, see [8], [4], [16], [11],
or [7].

The point of departure of our investigation is a lemma given in the book
“Chain Conditions in Topology” by W. W. Comfort and S. Negrepontis [4],
together with a question those authors posed. We give these now, paraphrasing
slightly to facilitate the present exposition.

Lemma 1.2. [4, 10.3] Let ω ≤ κ ≤ α, {Xi : i ∈ I} be a set of nonempty
topological spaces, and Y be a subspace of (XI )κ such that πJ [Y ] = XJ for
every nonempty J ⊂ I with |J| < α. Let also Z be a topological space and f
be a continuous function from Y to Z such that f depends on < α coordinates.
Then there is continuous f : (XI )κ → Z such that f ⊂ f.

Question 1.3. [4, p. 235] Let ω ≤ κ ≤ α, {Xi : i ∈ I} be a set of nonempty
topological spaces, and Y be a dense subspace of (XI )κ. Let Z be a space
such that πJ [Y ] is C(Z)-embedded in XJ for every nonempty J ∈ [I]

<α. If



On the continuity of factorizations 265

f ∈ C(Y,Z) and there is J ∈ [I]<α such that f depends on J, must f extend
continuously over (XI )κ? What about the case κ = ω?

In fact, the present authors do not know the answer to Question 1.3 even if
all the spaces Xi are assumed metrizable.

Let us specialize to the case κ = ω. It is clear from Theorem 3.5(c) below
that if the function f in Lemma 1.2 depends on some J ∈ [I]<α then the
factorization fJ : XJ → Z is continuous. Then since πJ [Y ] = XJ , the function
f := fJ ◦ πJ is a continuous extension of f that depends on J. Thus Lemma
1.2 has this consequence.

Theorem 1.4. Let α ≥ ω, XI be a product space, and Y be a subspace of XI
such that πJ [Y ] = XJ for every nonempty J ∈ [I]

<α. Let also Z be a topological
space and f ∈ C(Y,Z) depend on some J ∈ [I]<α. Then fJ is continuous and

f := fJ ◦ πJ : XI → Z is a continuous extension of f that depends on J.

Questions (a) through (e) of the Abstract are subsidiary to a more com-
pelling very general question: “When is a continuous function defined on a
dense subset of a product space continuously extendable over the full prod-
uct?” This question and question (a) of our Abstract have generated a huge
literature. Among the works in this vein, representing a variety of approaches,
we mention these familiar papers: H. Corson [5], R. Engelking [6], I. Glicksberg
[9], M. Hušek [12], [13], [14], A. Mishchenko [18], N. Noble [19], N. Noble and
M. Ulmer [20], M. Ulmer [22], [23]. The textbooks [7] and [4] strive for com-
prehensive bibliographies. As is indicated in [4], many of the published results
responding positively to these questions generalize to product spaces with the
κ-box topology.

In contrast, questions (b) through (e) of the Abstract have been almost to-
tally ignored in the literature. In this paper, always with f ∈ C(Y,Z) and
usually with Y dense in XI , we study some of these questions and their re-
lation to Question 1.3. In Section 2 we give some examples of discontinuous
factorizations fJ ; in Section 3 we give some sufficient conditions for the exis-
tence of continuous factorizations; in Section 4 we study when the existence of
a factorization of a function defined on a dense subspace of a product space
implies the existence of a factorization of its continuous extension to the full
product; in Section 5 we give some sufficient conditions for a positive answer
to Question 1.3; and in Section 6 we pose a question related to Question 1.3.

2. Discontinuous factorizations: some examples

In this section, responding to question (b) of the Abstract, we give examples
showing that a factorization of a continuous function defined on a dense subset
of a product space need not be continuous (2.3, 2.8, 2.9); and when it is con-
tinuous the initial function may (2.8) or may not (2.13) extend continuously
over the full product. Concerning question (e) of the Abstract, 2.3 and 2.9 give
examples of a function f ∈ C(Y,Z) with an extension f ∈ C(XI,Z) such that,



266 W. W. Comfort, I. S. Gotchev and L. Recoder-Núñez

for certain J ⊆ I, f does and f does not depend on J. As to question (d), the
answer is “Always” (4.6).

We use in what follows the familiar fact (see for example [7, 2.3.15] or [3,
3.18]) that for κ ≤ c the product of κ-many separable spaces is separable. For
convenience we give statements with each Xi = [0, 1], but routine generaliza-
tions (for example, with each Xi = R) are clearly valid.

Lemma 2.1. Let I be an index set with 0 < |I| ≤ c and XI = [0, 1]
I . Let

D = {x(n) : n < ω} be a (countable) dense subset of XI , and for n < ω let
y(n) ∈ XI satisfy |y(n)i − x(n)i| <

1
n+1

for each i ∈ I. Then E := {y(n) : n <

ω} is dense in XI .

Proof. Each nonempty open U ⊆ XI contains a (basic) set of the form

N(p,F,ǫ) :=
∏

i∈F

(pi − ǫ,pi + ǫ) × [0, 1]
I\F

with p ∈ XI , F ∈ [I]
<ω and ǫ > 0. The open set V := N(p,F, ǫ

2
) satisfies

|V ∩ D| = ω, and with n chosen so that 1
n+1

< ǫ
2

and x(n) ∈ V we have

y(n) ∈ N(p,F,ǫ) ∩ E ⊆ U ∩ E. �

Lemma 2.2. Let I be an index set with 0 < |I| ≤ c and XI = [0, 1]
I . There

is a countable dense subset E of XI such that for each i ∈ I the restriction
πi|E : E → [0, 1]i = [0, 1] is an injection.

Proof. Let D = {x(n) : n < ω} be dense in XI . Define y(0)i = x(0)i for
each i, and if y(n)i has been defined for all n < m choose y(m)i ∈ [0, 1] so
that y(m)i /∈ {y(0)i, . . . ,y(m − 1)i} and |y(m)i − x(m)i| <

1
m+1

. Then by

Lemma 2.1 the set E := {y(n) : n < ω} is as required. �

Example 2.3. There is a product space XI , a (countable) dense subspace
Y = {y(n) : n < ω} ⊆ XI , f ∈ C(Y ), and an index j ∈ I, such that

(i) the function fj : πj [Y ] → R given by fj (y(n)j ) := f(y(n)) is well-
defined, and

(ii) fj is not continuous on πj [Y ].

One may arrange in addition that f : Y → R extends continuously over XI .

Proof. Let each Xi = R or Xi = [0, 1] with |I| = c, and set XI :=
∏

i∈I Xi
and Y := E = {y(n) : n < ω} as in Lemma 2.2. Choose and fix two different
coordinates i,j ∈ I and define f : Y → R by f := πi|Y . Note that f extends
continuously over XI . Note also that the function fj : πj [Y ] → R given by
fj(y(n)j ) := f(y(n)) = y(n)i is well-defined, since if y(n),y(m) ∈ Y with
y(n)j = y(m)j then m = n. Now choose distinct x,y ∈ Y and let p = xi and
q = yj. Thus fj (q) 6= p. Let also nk be a sequence such that y(nk)i → p,
y(nk)j → q. Then fj (y(nk)j ) = f(y(nk)) = y(nk)i → p 6= fj(q), so fj is not
continuous. �

Theorem 2.4. Let I be an index set with 0 < |I| ≤ c and XI = [0, 1]
I .

There is a dense subspace Y of XI such that for each i ∈ I the restriction
πi|Y : Y ։ [0, 1]i = [0, 1] is a bijection onto [0, 1].



On the continuity of factorizations 267

Proof. Begin with E = {y(n) : n < ω} as in Lemma 2.2 and for i ∈ I let
{y(η)i : ω ≤ η < c} be a faithful enumeration of the set [0, 1]\πi[E]. Then
y(η) ∈ XI and the set Y := {y(η) : η < c} = E ∪ {y(η) : ω ≤ η < c} is as
required. �

Remark 2.5. In Example 2.3 the set πj [Y ] is a countable, dense subspace of
Rj = R or of [0, 1]j = [0, 1], hence is not C

∗-embedded. If we take I, XI ,
and Y as in Theorem 2.4 with I = {i,j}, |Y | = c, XI = R

2 and with both
πj and πi surjections from Y onto Xj = R and Xi = R, respectively, then we
can arrange the essential features of that argument. Note that the function
fj : πj [Y ] → R given by fj(y(η)j ) := f(y(η)) = y(η)i is still well-defined and
discontinuous since its restriction to the countable subspace E is discontinuous,
though fj ◦ πj = f with f(y(η)) = y(η)i is continuous on Y and as before, f
extends continuously over all of R2.

In Example 2.3 the function f depends on the set {i} and also depends on
the set {j} but the function fi is continuous while the function fj is not. The
following proposition shows, more generally, that if (continuous) f : Y → Z
depends on nonempty disjoint sets J1,J2 ⊂ J with fJ1 continuous, then either
f is a constant function or fJ2 is nowhere continuous.

Proposition 2.6. Let XI be a product space, Z be a Hausdorff space, and Y
be a dense subspace of XI . Let also J1 and J2 be nonempty disjoint subsets of
I, f : Y → Z be a non-constant (continuous) function that depends on J1 and
J2, and fJ1 is continuous. Then fJ2 is discontinuous at every point of πJ2 [Y ].

Proof. Let y ∈ Y . We shall show that fJ2 is discontinuous at yJ2 . Let
fJ2 (yJ2 ) = z1, hence f(y) = z1 and fJ1 (yJ1 ) = z1. Since the function f is
not constant there is x ∈ Y such that f(x) 6= z1. Let f(x) = z2, hence
fJ1 (xJ1 ) = z2. Since Z is a Hausdorff space we can find two disjoint open sets
U1 and U2 in Z such that z1 ∈ U1 and z2 ∈ U2. The function fJ1 is continuous
at xJ1 . Therefore there is a basic open neighborhood VJ1 of xJ1 in XJ1 such
that fJ1 [VJ1 ∩πJ1 [Y ]] ⊂ U2. Now, assume that there exists a basic open neigh-
borhood VJ2 of yJ2 in XJ2 such that fJ2[VJ2 ∩ πJ2 [Y ]] ⊂ U1. Since Y is dense
in XI and J1 ∩ J2 = ∅ there is t ∈ Y such that t ∈ π

−1
J1

[VJ1 ] ∩ π
−1
J2

[VJ2 ], hence
tJ1 ∈ VJ1 ∩ πJ1 [Y ] and tJ2 ∈ VJ2 ∩ πJ2 [Y ]. Then f(t) = fJ1 (tJ1 ) = fJ2 (tJ2 ), so
f(t) ∈ U1 ∩ U2 = ∅, a contradiction. �

Corollary 2.7. Let XI be a product space, Z be a Hausdorff space, and Y be
a dense subspace of XI . Let also J1 and J2 be nonempty disjoint subsets of I,
f : Y → Z be a (continuous) function that depends on J1 and J2, and fJ1 and
fJ2 are continuous. Then f is constant.

Example 2.3 suggests the speculation that if some f ∈ C(Y,Z) depends on
a set J with fJ discontinuous, and if f extends continuously over XI , then one
can find another set J1 ⊂ I with |J1| = |J| such that f depends on J1 and fJ1
is continuous. We show in Corollary 2.9 below that this can fail: there exist
a cardinal number α ≥ ω, a product space XI , a Hausdorff space Z, a dense



268 W. W. Comfort, I. S. Gotchev and L. Recoder-Núñez

subspace Y of XI , and a continuous function f : Y → Z such that for every
J ∈ [I]<α, f depends on J but fJ is not continuous, and the function f can be
extended to a continuous function f on XI . (There is, however, a fragment of
Question 1.3—logically, an equivalent formulation—which survives in the face
of Corollary 2.9. We give the statement in 6.1 below.)

In what follows we take I, XI , and Y as in Theorem 2.4 with |I| ≥ ω,
|Y | = c, and with each πi|Y : Y ։ [0, 1] a bijection onto [0, 1]; the indexing
Y = {y(η) : η < c} plays no further role. We note that for every set Z, every
function f : Y → Z depends (vacuously) on each set {i} with i ∈ I. Thus, for
every f : Y → Z and ∅ 6= J ⊆ I the function fJ : πJ [Y ] → Z is well-defined
by the rule fJ (pJ ) := f(p) (p ∈ Y ).

Theorem 2.8. Given Y ⊆ XI = [0, 1]
I as above, let C = {in : n < ω} ∈

[I]ω (faithfully indexed) and define g : XI → [0, 1] by g(x) :=
∑

n<ω
xin
2n

=
∑

n<ω
πin (x)

2n
. Then for ∅ 6= J ⊂ I the function gJ : πJ [Y ] → [0, 1] is continu-

ous if and only if C ⊆ J.

Proof. (We note that the series defining g converges uniformly on XI , so g :
XI → [0, 1] is continuous.)

If. Let pJ ∈ πJ [Y ] ⊆ XJ with p ∈ Y , and let p(λ)J be a net in πJ [Y ] (with
p(λ) ∈ Y ) such that p(λ)J → pJ . Then p(λ)i → pi for each i = in ∈ C, so
g(p(λ)) → g(p), i.e., gJ (p(λ)J ) → gJ (pJ ).

Only if. We show (when C ⊆ J fails) that gJ is continuous at no point
pJ ∈ πJ [Y ].

Fix i = in ∈ C\J and p ∈ Y and define q ∈ XI by qi := pi if i 6= i ∈ I,
|qin − pin | =

1
4
. Since qi = pi for all i 6= in, in particular for all i ∈ C\{in},

we have |g(p) − g(q)| = 1
4
· 1

2n
. There is a net p(λ) in Y such that p(λ) → q.

Since pJ = qJ we have p(λ)J → qJ = pJ ∈ πJ [Y ], but from the continuity of g
on XI we have gJ (p(λ)J ) = g(p(λ)) → g(q) 6= g(p) = gJ (pJ )). Thus gJ is not
continuous on πJ [Y ]. �

Corollary 2.9. For ω ≤ κ ≤ c there are a dense subset Y of [0, 1]κ and
continuous f : Y → [0, 1] such that

(a) f depends on every nonempty J ⊆ κ;
(b) each restricted projection πη|Y : Y ։ [0, 1]η = [0, 1] (η < κ) is a bijection

onto [0, 1];
(c) there is no J ∈ [κ]<ω such that fJ : πJ [Y ] → [0, 1] is continuous; and
(d) f extends to a continuous function f : [0, 1]κ → [0, 1].

Proof. From Theorem 2.8, taking f := g|Y . �

The examples just given show that a function g ∈ C(XI,Z) may fail to
depend on a set J, even when f := g|Y : Y → Z does depend on J. We discuss
this in greater detail in Section 4.



On the continuity of factorizations 269

Discussion 2.10. We draw the reader’s attention to two hypotheses in Ques-
tion 1.3.

(a) there is J ∈ [I]<α such that f depends on J; and
(b) πJ [Y ] is C(Z)-embedded in XJ for every nonempty J ∈ [I]

<α.
We end this section with examples showing that if either (a) or (b) is not

satisfied then the resulting weaker questions can be answered in the negative.
Again we specify to the case κ = ω.

Example 2.11. There are α ≥ ω, a Tychonoff product space XI , a dense
subspace Y ⊂ XI , and f ∈ C(Y ) such that πJ [Y ] is C-embedded in XJ for all
nonempty J ∈ [I]<α, but f has no continuous extension from XI to R and f
does not depend on any proper nonempty subset J ∈ [I]<α.

Proof. We are aware of two relevant constructions from the literature.
(1) [the case α = ω] Let N denote the countably infinite discrete space. It

is shown in [1] that there is a sequence {Yk : k < ω} of spaces, with in each
case N ⊆ Yk ⊆ β(N) := Xk, such that Y := Πk<ω Yk is not pseudocompact
but YJ = Πk∈J Yk is pseudocompact whenever J ∈ [ω]

<ω. It follows from
Glicksberg’s theorem [9] that β(YJ ) = Πk∈J Xk for each J ∈ [ω]

<ω (so πJ [Y ] =
YJ is C

∗-embedded in XJ ), but the relation β(Y ) = Πk<ω Xk fails (so some
f ∈ C∗(Y ) has no continuous extension from XI = Πk<ω Xk to R).

If f depends on some nonempty J ∈ [ω]<ω then (since Y is a product space)
Theorem 3.5(a) shows fJ ∈ C

∗(YJ ). Then fJ extends to fJ ∈ C
∗(XJ ) and we

have f ⊆ fJ ◦ πJ ∈ C
∗(XI ), a contradiction.

(2) [arbitrary α ≥ ω] Ulmer [22], [23] has given many examples, enhanced
and extended in our work [2], of a Tychonoff product space XI and a Σα-
product subspace Y ⊂ XI which is not C-embedded in XI (so some f ∈ C(Y )
has no continuous extension from XI to R).

If f depends on some nonempty J ∈ [I]<α then fJ ∈ C(πJ [Y ]) since Y is
a Σα-space (see Theorem 3.5(b)). πJ [Y ] is trivially C-embedded in XJ since

πJ [Y ] = XJ , and we have the contradiction f ⊆ fJ ◦ πJ ∈ C(XI ). �

Example 2.12. There are a Tychonoff product space XI = Πi∈I Xi, a dense
subspace Y ⊂ XI , a function f ∈ C(Y ), and J ∈ [I]

<ω such that

(a) f depends on J,
(b) πJ [Y ] is C-embedded in XJ , and
(c) f does not extend continuously over XI .

Proof. In [2, 3.2] we have shown, extending arguments introduced by Ulmer
[22], [23], that there are a Tychonoff space XI′ = Πi∈I′ Xi with |Xi| = |I

′| = ω
for each i ∈ I′, q′ ∈ XI′ , and a continuous function f

′ : XI′\{q
′} → [0, 1]

which does not extend continuously over XI′ . We arrange the notation so that
no symbol in I′ is named 0, and we set I := I′ ∪ {0} and X0 := [0, 1]. Since
{q′} × X0 is closed and nowhere dense in XI , there is a countable dense subset
{x(n) : n < ω} of XI which misses {q

′} × X0, and a routine modification of
the argument in Lemmas 2.1 and 2.2 gives a dense set E = {y(n) : n < ω},
also missing {q′} × X0, such that the restricted projection π0|E : E → X0 is



270 W. W. Comfort, I. S. Gotchev and L. Recoder-Núñez

an injection. (Arguing recursively one lets y(m)I′ = x(m)I′ and one chooses
(distinct) points y(m)0 ∈ [0, 1] such that |y(m)0 − x(m)0| <

1
m+1

.) Since

|(XI′\{q
′})\πI′ [E]| = |[0, 1]\π0[E]| = c, there is a set Y such that E ⊆ Y ⊆ XI

and πI′ [Y ] = XI′\{q
′} and π0|Y is a bijection onto X0. We define f : Y → [0, 1]

by f(r,x′) := f′(x′) for (r,x′) ∈ Y ⊆ X0 × (XI′\{q
′}) ⊆ X0 × XI′ = XI .

To see that f is continuous on Y it is enough to note that if y(λ) is a net in
Y such that y(λ) → y = (r,x′) ∈ Y then y(λ)I′ → x

′ with y(λ)I′ ∈ XI′\{q
′}

and hence f(y(λ)) = f′(y(λ)I′ ) → f
′(x′) = f(y). Clearly f depends on the

coordinate 0 ∈ I, i.e., f depends on J := {0} ∈ [I]<ω, and π0[Y ] is trivially
C-embedded in X0 since π0[Y ] = X0. Also f depends on I

′, so there can be no

continuous extension f of f over XI : According to the implication (i) ⇒ (iv)
of Theorem 4.6 below (with our f′, I′, and f playing the role there of f, J,
and g, respectively), the existence of such continuous f would yield f′ ⊆ fI′ ∈
C(XI′, [0, 1]), a contradiction. �

Example 2.13. There is a Tychonoff product space XI = Πi∈I Xi such that
for every nonempty J ∈ [I]≤ω there are a dense subspace Y ⊂ XI and a
function f ∈ C(Y ) such that

(a) f depends on J,
(b) fJ is continuous,
(c) fJ does not extend continuously over XJ , and
(d) f does not extend continuously over XI .

Proof. Let {Xi : i ∈ I} be a set of metrizable spaces without isolated points,
let J ⊆ I satisfy 0 < |J| ≤ ω, fix p ∈ XJ and set D := XJ \{p}. Some g ∈
C(D, [0, 1]) admits no continuous extension over XJ , and then Y := π

−1
J (D)

and f := g ◦ πJ |Y are as required (with fJ = g and πJ [Y ] = D). �

3. Existence of continuous factorizations

In this section we give some conditions that imply the continuity of functions
of the form fJ . We begin with the following observation.

Lemma 3.1. Let X, Y , and Z be spaces and f : X → Y , g : X → Z, and
h : Z → Y be functions such that f = h ◦ g. If f is continuous and g is open,
then h|g[X] is continuous.

Proof. Take U an open set in Y . Then (h|g[X])
−1[U] = g[f−1[U]]. �

Theorem 3.2. Let XI be a product space, J be a nonempty subset of I, and Y
be a nonempty subspace of XI such that πJ [U ∩ Y ] = πJ [U] ∩ πJ [Y ] for every
basic open set U of XI . Let also Z be a space and f ∈ C(Y,Z) depend on J.
Then fJ : πJ [Y ] → Z is continuous.

Proof. Since f = fJ ◦ πJ |Y with f continuous and πJ |Y open, the continuity
of fJ follows from Lemma 3.1. �



On the continuity of factorizations 271

Lemma 3.3. Let XI be a product space, J be a nonempty proper subset of I,
and Y be a subset of XI . The set πI\J [π

−1
J (πJ (y)) ∩ Y ] is dense in XI\J for

every y ∈ Y if and only if πJ [U] ∩ πJ [Y ] = πJ [U ∩ Y ] for every basic open set
U in XI .

Proof. Let the set πI\J [π
−1
J (πJ (y)) ∩ Y ] be dense in XI\J for every y ∈ Y and

U be a basic open set in XI . We shall prove that πJ [U] ∩ πJ [Y ] = πJ [U ∩ Y ].
Let t ∈ πJ [U] ∩ πJ [Y ] and let x ∈ Y be such that xJ = t and xI\J ∈ πI\J [U] ∩

πI\J [Y ]. (Such a point x exists since πI\J [π
−1
J (t) ∩ Y ] is dense in XI\J .) Then

x ∈ U ∩ Y , hence t ∈ πJ [U ∩ Y ]. Therefore πJ [U] ∩ πJ [Y ] = πJ [U ∩ Y ].
Now, let πJ [U]∩πJ [Y ] = πJ [U∩Y ] for every basic open set U in XI . We shall

prove that πI\J [π
−1
J

(πJ (y)) ∩ Y ] is dense in XI\J for every y ∈ Y . Let y ∈ Y

and V be a basic open set in XI\J . Then W = π
−1
I\J

[V ] is a basic open set in XI
and πJ (y) ∈ πJ [W ]∩πJ [Y ] = πJ [W ∩Y ]. Therefore there exists x ∈ W ∩Y such
that xJ = yJ and xI\J ∈ πI\J [W ] = V , hence xI\J ∈ πI\J [π

−1
J

(πJ (y))∩Y ]∩V .

Thus πI\J [π
−1
J

(πJ (y)) ∩ Y ] is dense in XI\J . �

Remark 3.4. (a) It is clear that when XI , Y and J satisfy the (equivalent)
conditions of Lemma 3.3, the function πJ |Y is an open map. It is useful to
note that the converse can fail. For an example, let X0 and X1 be nonempty
T1-spaces with each |Xi| > 1, fix (x0,x1) ∈ X0 ×X1 and set Y := ((X0\{x0})×
X1)∪{(x0,x1)}. Then π0|Y is an open map, but π1[π

−1
0 (π0(x0,x1))∩Y ], which

is the singleton set {π1(x0,x1)} = {x1}, is not dense in X1. (Alternatively:
with U := X0 × (X1\{x1}) we have x0 ∈ (π0[U] ∩ π0[Y ])\π0[U ∩ Y ].)

(b) In a trivial way, using Lemma 3.1, fJ will be continuous provided Y is
a nonempty open subspace of a product space XI and f ∈ C(XI,Z) depends
on J ⊂ I.

Theorem 3.5. Let XI be a product space, J be a nonempty proper subset of
I, α = |J|+, and Y be a nonempty subspace of XI . Let also Z be a space and
f ∈ C(Y,Z) depend on J. Then fJ : πJ [Y ] → Z is a continuous function if

(a) Y = XI ; or
(b) [y ∈ Y ] ⇒ Σα(y) ⊆ Y ; or
(c) [J′ ⊆ I, |J′| ≤ |J|] ⇒ πJ′ [Y ] = XJ′ ; or
(d) [J′ ⊆ I,J′ = J ∪ F with |F | < ω] ⇒ πJ′ [Y ] = XJ′ ; or
(e) π−1

J
[πJ [Y ]] = Y ; or

(f) πI\J [π
−1
J (πJ (y)) ∩ Y ] is dense in XI\J for every y ∈ Y ; or

(g) πJ [U] ∩ πJ [Y ] = πJ [U ∩ Y ] for every basic open set U in XI ; or
(h) πJ [Y ] × {yI\J } ⊆ Y for some y ∈ Y .

Proof. Clearly (a) ⇒ (b), (b) ⇒ (c), and (c) ⇒ (d). To see that (d) ⇒ (g), let
U be a basic open set in XI and xJ ∈ πJ [U] ∩ πJ [Y ] with x ∈ U, and define
J′ := J ∪ R(U); then since πJ′ [Y ] = XJ′ there is y ∈ Y such that yJ′ = xJ′ ,
so xJ = yJ ∈ πJ [U ∩ Y ]. We conclude that πJ [U] ∩ πJ [Y ] = πJ [U ∩ Y ]. If
(e) holds then πI\J [Y ] = XI\J , so (f) holds, and if (f) holds then (g) holds
by Lemma 3.3. Thus by Theorem 3.2 the function fJ is continuous under any



272 W. W. Comfort, I. S. Gotchev and L. Recoder-Núñez

of the conditions (a), (b), (c), (d), (e), (f), and (g). If (h) holds and h is the
natural homeomorphism from πJ [Y ] onto Y

′ := πJ [Y ] × {yI\J }, then f|Y ′ is

defined and fJ = f|Y ′ ◦ h. �

Remark 3.6. Theorem 3.5 serves to show that any dense subspace Y of a
product space XI for which some f ∈ C(Y,Z) depends on J ⊆ I with fJ dis-
continuous must have properties in common with the spaces from our examples
in Section 2. Thus if the answer to Question 1.3 is “No” then the witnessing ex-
ample must have some of the properties of the dense spaces Y in our examples
in Section 2.

4. Factorizations and continuous extensions

Let g ∈ C(XI,Z), Y be a dense subset of XI , f := g|Y , and J ⊂ I. We
know from Example 2.3 and Corollary 2.9 that g may fail to depend on J even
if f does depend on J. In this section we give additional conditions sufficient
to ensure that this counterintuitive phenomenon cannot occur.

Theorem 4.1. Let XI be a product space, J be a nonempty proper subset of
I, and Y be a nonempty subspace of XI such that πJ [Y ] × {yI\J } ⊆ Y for

some y ∈ Y . Let also Z be a space and g ∈ C(XI,Z) be such that the function
f := g|Y depends on J. Then fJ is continuous and has a continuous extension
fJ : XJ → Z. Therefore f has a continuous extension f : XI → Z that depends
on J.

Proof. The function fJ is continuous according to Theorem 3.5(h). Let T =
XJ × {yI\J }. If h := g|T then h : T → Z is continuous and depends on

J. Therefore, according to Theorem 3.5(h) again, the function hJ : XJ → Z

is continuous. Since hJ |πJ [Y ] = fJ the function fJ := hJ is a continuous

extension of fJ . Then the function f := fJ ◦ πJ is a continuous extension of f
that depends on J, as required. �

Corollary 4.2. Let XI be a product space, J be a nonempty proper subset of
I, and Y be a dense subspace of XI such that πJ [Y ] × {yI\J } ⊆ Y for some

y ∈ Y . Let also Z be a Hausdorff space and g ∈ C(XI,Z) be such that the
function f := g|Y depends on J. Then g depends on J.

Proof. The domain of agreement of two continuous functions from a fixed space
to a Hausdorff space is closed [7, 2.1.9], so with Z Hausdorff we have f = g in
Theorem 4.1. �

The following two corollaries are immediate from Corollary 4.2.

Corollary 4.3. Let α ≥ ω, XI be a product space, Z be a Hausdorff space,
g ∈ C(XI,Z), J be a nonempty proper subset of I, and Y ⊆ XI be a Σα-space.
Then g depends on J if and only if f := g|Y depends on J.



On the continuity of factorizations 273

Corollary 4.4. Let XI be a product space, Z be a Hausdorff space, g ∈
C(XI,Z), J be a nonempty proper subset of I, and Y be a dense subspace
of XI such that Y = πJ [Y ] × πI\J [Y ]. Then g depends on J if and only if
f := g|Y depends on J.

The following theorem is a special case of Corollary 4.4. The countable case
(|J| ≤ ω) is mentioned without proof by N. Noble and M. Ulmer in the proof
of Proposition 2.1 in [20].

Theorem 4.5. Let XI be a product space, Z be a Hausdorff space, g ∈
C(XI,Z), Yi be a dense subspace of Xi for every i ∈ I, Y =

∏
i∈I Yi, and

J be a nonempty proper subset of I. Then g depends on J if and only if
f := g|Y depends on J.

The following theorem contains conditions equivalent to the continuity of a
given factorization.

Theorem 4.6. Let XI be a product space, Z be a Hausdorff space, g ∈
C(XI,Z), Y be a dense subspace of XI , f := g|Y , and J be a nonempty subset
of I. Then the following are equivalent.

(i) f depends on J and fJ : πJ [Y ] → Z is continuous.
(ii) h := g|

π
−1

J
[πJ [Y ]]

depends on J.

(iii) g depends on J.
(iv) f depends on J, fJ is continuous, and fJ has a continuous extension

fJ : XJ → Z.

Proof. (i) ⇒ (ii). Let x,y ∈ π−1
J

[πJ [Y ]] and z ∈ Y be such that xJ = yJ = zJ .
Then there exists a net (xα) ⊂ Y with limit x. Since g is continuous we have
(g(xα)) → g(x). Therefore (fJ ((xα)J )) → g(x) for g(xα) = fJ ((xα)J ) for each
α, and since fJ is continuous (fJ ((xα)J )) → fJ (xJ ). Thus g(x) = fJ (xJ ) =
fJ (zJ ) = g(z). Similarly we have g(y) = g(z). Therefore g(x) = g(y).

(ii) ⇒ (iii). Let x,y ∈ XI be such that xJ = yJ . If xJ ∈ πJ [Y ], then
x,y ∈ π−1

J
[πJ [Y ]] and therefore g(x) = g(y) for g|π−1

J
[πJ [Y ]]

depends on J. Now,

let xJ /∈ πJ [Y ]; then x,y /∈ π
−1
J

[πJ [Y ]]. Since πJ
−1[πJ [Y ]] is dense in XI there

exists a net (xα) ⊂ π
−1
J [πJ [Y ]] such that (xα) → x. Then (g(xα)) → g(x) for g

is continuous. For each α, we define wα(i) = xα(i) for all i ∈ J and wα(i) = y(i)
for all i ∈ I \J. It is clear that the net (wα) → y. Then (g(wα)) → g(y) for
g is continuous. Also, wα,xα ∈ π

−1
J

[πJ [Y ]] and (wα)J = (xα)J for every α
and since h = g|

π
−1

J
[πJ [Y ]]

depends on J we have g(wα) = g(xα) for every α.

Therefore g(x) = g(y). Thus g depends on J.
(iii) ⇒ (iv). If g depends on J then gJ is a continuous function which extends

fJ .
(iv) ⇒ (i). Obvious. �

The following theorem is immediate from Theorem 4.6.



274 W. W. Comfort, I. S. Gotchev and L. Recoder-Núñez

Theorem 4.7. Let XI be a product space, Z be a Hausdorff space, g ∈
C(XI,Z), and J be a nonempty subset of I. The function g depends on J
if and only if there exists a dense subspace Y of XI such that f := g|Y depends
on J and fJ is continuous.

As an illustrative application of some of our results we provide now in Theo-
rem 4.10 a proof of a generalization of a classical theorem of A. M. Gleason (see
[21, p. 401], [15], or [6]). More general versions of that theorem, with different
proofs, can be found in [18], [12], or [14].

Lemma 4.8. Let α ≥ ω, XI be a product space, Y ⊆ XI be a dense subset
with |Y | ≤ α, Z be a T1-space such that χ(Z) ≤ α, and f ∈ C(Y,Z). Then
there exists J ∈ [I]≤α such that f depends on J and fJ is continuous.

Proof. For y ∈ Y let {U(y)a : a ∈ A} be a local base at f(y) in Z with |A| ≤ α.
Since the function f is continuous at y, for every U(y)a, a ∈ A, we can find
a basic open neighborhood V (y)a of y in XI such that f[V (y)a ∩ Y ] ⊂ U(y)a.
Let J := ∪y∈Y (∪a∈A R(V (y)a)). If two points x,y ∈ Y are such that xJ = yJ
then f(x) = f(y), hence f depends on J.

Now, let w ∈ πJ [Y ] and y ∈ Y be such that w = yJ . Let also U be
an open neighborhood of fJ (w) = f(y) in Z. Then there exists b ∈ A such
that f[V (y)b ∩ Y ] ⊂ U and since R(V (y)b) ⊂ J we have πJ [V (y)b] ∩ πJ [Y ] =
πJ [V (y)b ∩ Y ]. Therefore fJ [πJ [V (y)b] ∩ πJ [Y ]] ⊂ U. We conclude that fJ is
continuous at w. �

Lemma 4.8 can be “localized” as follows.

Corollary 4.9. Let XI be a product space, Z be a T1-space, Y ⊆ XI , J be an
infinite subset of I, f ∈ C(Y,Z) depend on J, y ∈ Y , and χ(f(y),Z) ≤ |J|.
Then there exists Jy ⊆ I such that |Jy| = |J|, f depends on Jy, and fJy is
continuous at yJy .

Proof. Let {Ua : a ∈ A} be a local base at f(y) in Z with |A| ≤ |J|. Since
the function f is continuous at y, for every Ua, a ∈ A, we can find a basic
open neighborhood Va of y in XI such that f[Va ∩ Y ] ⊂ Ua. Let Jy = J ∪
(∪a∈A R(Va)). Then |Jy| = |J|, f depends on Jy, and fJy is continuous at
yJy . �

Theorem 4.10. Let α ≥ ω, {Xi : i ∈ I} be a set of spaces with d(Xi) ≤ α for
each i ∈ I, Z be a Hausdorff space with χ(Z) ≤ α, and g ∈ C(XI,Z). Then g
depends on ≤ α coordinates.

Proof. We assume without loss of generality, replacing Z by g[XI ] if necessary,
that g : XI → Z is a surjection. Since d(Xi) ≤ α for each i ∈ I the cellularity
c(XI ) of XI is ≤ α (see [7, 2.3.17], [4, 3.28], [16, 5.6]). Thus, c(Z) ≤ α and
since χ(Z) ≤ α and Z is Hausdorff we have |Z| ≤ 2c(Z)χ(Z) = 2α (see [16,
2.15(b)]), [11]).

For z ∈ Z let {U(z)a : a ∈ A} be a local base at z in Z with |A| ≤ α. The
function g is continuous, hence the set g−1[U(z)a] is open for every z ∈ Z and



On the continuity of factorizations 275

every a ∈ A. Since XI is a product of spaces with d(Xi) ≤ α for each i ∈ I,
the closure of every open set depends on ≤ α many coordinates (see [4, 10.13]).
For z ∈ Z and a ∈ A, let J(z,a) ∈ [I]≤α be a nonempty set such that the set

g−1[U(z)a] depends on it and let K := ∪z∈Z (∪a∈AJ(z,a)). If x,y ∈ XI are

such that xK = yK and g(x) = z then y ∈ g−1[U(z)a] ⊆ g
−1[U(z)a] for every

a ∈ A and since Z is Hausdorff {z} = ∩a∈AU(z)a, thus y ∈ g
−1(z). Therefore

g(x) = g(y), hence g depends on K. Thus, gK : XK → Z is continuous and
since |K| ≤ 2α it follows from Hewitt-Marczewski-Pondiczery theorem (see [7,
2.3.15]) that XK contains a dense set Y ∈ [XK ]

≤α. Then by Lemma 4.8 (with
XK now in place of XI ) there exists J ∈ [K]

≤α such that gK depends on J.
Therefore g depends on J. �

Remark 4.11. As it is clear from the proof of Theorem 4.10, the hypothesis
χ(Z) ≤ α there may be relaxed to the condition 2χ(Z) ≤ 2α.

Since χ(Z) ≤ w(Z) for every space Z, Theorem 4.10 gives a proof of Theorem
10.14 in [4]:

Corollary 4.12. Let α ≥ ω, {Xi : i ∈ I} be a set of spaces with d(Xi) ≤ α
for i ∈ I, Z be a Hausdorff space with w(Z) ≤ α, and g ∈ C(XI,Z). Then g
depends on ≤ α coordinates.

The foregoing results afford several conditions which ensure that (under
suitable hypotheses) a function g : XI → Z depends on a set J ⊆ I iff the
restricted function g|Y does so. We record two instances of particular interest.

Theorem 4.13. Let XI be a product space, Z be a Hausdorff space, g ∈
C(XI,Z), J be a nonempty proper subset of I, and Y be a dense subset of XI .
If either

(a) [J′ ⊆ I, |J′| ≤ |J|] ⇒ πJ′ [Y ] = XJ′ , or
(b) πI\J [π

−1
J (πJ (y)) ∩ Y ] is dense in XI\J for every y ∈ Y ,

then g depends on J iff f := g|Y depends on J.

Proof. Surely if g depends on J then f = g|Y depends on J. For the reverse
implications, it is enough to note from Theorem 3.5 that fJ is continuous, so
the implication (i) ⇒ (iii) of Theorem 4.6 applies. �

5. On the Comfort-Negrepontis question (Question 1.3)

In this section we give some sufficient conditions for a positive answer to the
Comfort–Negrepontis question. We begin with a workable equivalent condition
which in particular cases is readily verified (or, as in Section 2, refuted).

Theorem 5.1. Let α ≥ ω, XI be a product space, Z be a space, Y ⊆ XI be
a dense subspace such that πJ [Y ] is C(Z)-embedded in XJ for every nonempty
J ∈ [I]<α, and f ∈ C(Y,Z). The function f has a continuous extension
f : XI → Z that depends on < α coordinates if and only if there exists a
nonempty J ∈ [I]<α such that f depends on J and fJ is continuous.



276 W. W. Comfort, I. S. Gotchev and L. Recoder-Núñez

Proof. If f ∈ C(XI,Z) depends on some nonempty J ∈ [I]
<α then fJ is

continuous (see Theorem 3.2). Therefore fJ is continuous.
If f depends on J and fJ is continuous, then since πJ [Y ] is C(Z)-embedded

in XJ there is fJ such that fJ ⊆ fJ ∈ C(XJ,Z). Then f := fJ ◦ πJ is a
continuous extension of f and f depends on J. �

It is easily seen from the proof of the above theorem that if we need to
extend a particular function f that depends on a given nonempty set J ⊆ I
then we do not require the strong hypothesis that πJ [Y ] is C(Z)-embedded
in XJ . We need only that fJ is continuous and that it extends to continuous
fJ : XJ → Z. Therefore Theorem 5.1 could be restated as follows.

Theorem 5.2. Let XI be a product space, Z be a space, J be a nonempty
subset of I, Y ⊆ XI be a dense subspace, and f ∈ C(Y,Z) depend on J. The

function f has a continuous extension f : XI → Z that depends on J if and
only if fJ is continuous and has a continuous extension fJ : XJ → Z.

The next theorem generalizes Theorem 1.4 and gives conditions sufficient
that the answer to Question 1.3 is in the affirmative.

Theorem 5.3. Let XI be a product space, J be a nonempty proper subset of
I, α = |J|+, and Y be a nonempty subspace of XI . Let also Z be a space and
f ∈ C(Y,Z) depend on J. If

(a) Y = XI ; or
(b) [y ∈ Y ] ⇒ Σα(y) ⊆ Y ; or
(c) [J′ ⊆ I, |J′| ≤ |J|] ⇒ πJ′ [Y ] = XJ′ ; or
(d) [J′ ⊆ I,J′ = J ∪ F with |F | < ω] ⇒ πJ′ [Y ] = XJ′ ; or
(e) π−1J [πJ [Y ]] = Y and πJ [Y ] is C(Z)-embedded in XJ ; or

(f) πI\J [π
−1
J (πJ (y)) ∩ Y ] is dense in XI\J for every y ∈ Y and πJ [Y ] is

C(Z)-embedded in XJ ; or
(g) πJ [U] ∩ πJ [Y ] = πJ [U ∩ Y ] for every basic open set U in XI and πJ [Y ]

is C(Z)-embedded in XJ ; or
(h) πJ [Y ] × {yI\J } ⊆ Y for some y ∈ Y and πJ [Y ] is C(Z)-embedded in

XJ

then f has a continuous extension f : XI → Z that depends on J.

Proof. As in Theorem 3.5, clearly (a) ⇒ (b), (b) ⇒ (c), (c) ⇒ (d), and
(d) ⇒ (g). Also (e) ⇒ (f) and (f) ⇒ (g). If (g) holds then fJ in continu-
ous by Theorem 3.5; thus f := fJ ◦ πJ |Y is a continuous extension of f that
depends on J, so under any of the conditions (a), (b), (c), (d), (e), (f), and (g)
f has a continuous extension that depends on J. Similarly, if (h) holds then fJ
is continuous by Theorem 3.5; thus f has a continuous extension that depends
on J. �

A partial answer, in the positive, to Question 1.3 is given in Theorem 5.6.



On the continuity of factorizations 277

Lemma 5.4. Let X be a space, p ∈ X, and (Wn) be a sequence of open subsets

of X such that Wn+1 ⊆ W
X

n+1 ⊆ Wn and ∩n Wn = {p}. Then (Wn) is locally
finite at each point of X\{p}.

Proof. We are to show for p 6= x ∈ X that some neighborhood U of x meets Wn

for only finitely many n. Given x, it is enough to choose k such that x /∈ W
X

k

and to set U := X\W
X

k . �

We say as usual that a subset Y of a space X is sequentially closed if [yn ∈ Y
and yn → p ∈ X] ⇒ p ∈ Y , and X is a sequential space if every sequentially
closed subset of X is closed.

Theorem 5.5. Let X be a Tychonoff space with countable pseudocharacter and
let Y be a C∗-embedded subset of X. Then Y is sequentially closed in X.

Proof. Suppose there is a sequence yn ∈ Y such that yn → p ∈ X\Y , and let
(Un) be a (countable) local pseudobase at p in X.

Let Vn0 := U0 and choose yn0 ∈ Vn0 . Suppose that Vnk and ynk have
been defined. Let nk+1 > nk be such that ynk+1 ∈ Vnk+1 , where Vnk+1 is a

neighborhood in X of p such that Vnk+1 ⊆ V
X

nk+1
⊆ Vnk ∩Unk and ynk /∈ V

X

nk+1
.

With the sequences Vnk and ynk so defined, choose a function fk ∈ C(X, [0, 1])
such that fk ≡ 0 for k even, and for odd k, using the Tychonoff property of X,

such that 0 ≤ fk ≤ 1, fk(ynk ) = 1, and fk ≡ 0 on (X\Vnk ) ∪ V
X

nk+1
. Since the

sequence Vnk \V
X

nk+1
is pairwise disjoint, the function f := Σk fk : X → [0, 1]

is well-defined. According to Lemma 5.4, the sequence (Vnk ) is locally finite

at each point of X\{p}, so the sequence ((Vnk \V
X

nk+1
) ∩ Y ) is locally finite

in Y ; thus g := f|Y ∈ C
∗(Y ). For each neighborhood U of p in X there is

k such that yn2k,yn2k+1 ∈ U. We have yn2k+1 ∈ U ∩ Y with g(yn2k+1 ) = 1
and yn2k ∈ Vn2k ∩ Y with g(yn2k ) = 0. We conclude that g does not extend
continuously to p, a contradiction. �

Theorem 5.6. Let α ≤ ω1 and S be a class of Tychonoff spaces such that if
Sn ∈ S (n < ω, repetitions permitted) then Πn<ω Sn ∈ S and Πn<ω Sn is a
sequential space. Let also {Xi : i ∈ I} ⊆ S with ψ(Xi) ≤ ω for each i ∈ I, and
Y be a dense subspace of XI such that πJ [Y ] is C

∗-embedded in XJ for every
nonempty J ∈ [I]<α. If f ∈ C∗(Y ) and there is J ∈ [I]<α such that f depends
on J, then f extends continuously over XI .

Proof. By virtue of Lemma 1.2 (taking κ = ω and Z = R there), it suffices to
prove that πJ [Y ] = XJ for all J ∈ [I]

<α. Take J ∈ [I]<α. From ψ(Xi) ≤ ω (all
i ∈ J) and |J| ≤ ω it follows easily that ψ(XJ ) ≤ ω, so πJ [Y ] is sequentially
closed in XJ by Theorem 5.5 and hence closed in XJ , so πJ [Y ] = XJ . �

Remark 5.7. The class of all Tychonoff, first countable spaces of course sat-
isfies the hypotheses required of S in the statement of Theorem 5.6, but other
less familiar classes of spaces do so as well. For example, one may take for S
the class of all Tychonoff, bi-sequential spaces (see [17, Definition 3.D.1]) with



278 W. W. Comfort, I. S. Gotchev and L. Recoder-Núñez

countable pseudocharacter but not first countable. (It is known that every bi-
sequential space is sequential ([17]); the class of bi-sequential spaces is closed
under countable products ([17]); there exists a bi-sequential Tychonoff space
without countable pseudocharacter ([17, Example 10.4]); and there exists a bi-
sequential Tychonoff space with countable pseudocharacter which is not first
countable (see Example 5.8 below).)

We are indebted to Gary Gruenhage for the following example.

Example 5.8. Let K := [0, 1] be the interval [0, 1] with the usual topology.
For every rational q ∈ K, choose the constant sequence Sq := {q} and for every
irrational r ∈ K, choose a sequence Sr of rational numbers in K converging to
r. Define S := ∪x∈KSx and X := S ∪ {∞}. Let T be the smallest topology on
X such that:

(1) For every q ∈ S, {q} is T -open in X; and
(2) For every finite subset F of S and every finite subset G of [0, 1] the set

N(∞,F,G) := {∞} ∪ [S\(F ∪ ∪x∈GSx)] is a T -open neighborhood of ∞.
It follows from [10, Proposition 3.2] that X is bi-sequential. The family

{N(∞, ∅,{q}) : q ∈ S} is countable and {∞} = ∩q∈SN(∞, ∅,{q}). Hence
ψ(X) = ω. Now take a countable family {N(∞,Fn,Gn) : n < ω} of neighbor-
hoods of ∞. Choose an irrational r ∈ K not in ∪n<ωGn. Then N(∞, ∅,{r})
contains no N(∞,Fn,Gn), n < ω. Therefore X is not first countable. Since
each of the T -basic open sets specified in (1) and (2) is also T -closed, the
Hausdorff space (X,T ) is a Tychonoff space.

6. Concluding remarks

We remark that most of the results of the foregoing sections admit natural
generalizations to the κ-box topology. Since these are routine and their for-
mulation adds little to our broad understanding, we leave the details to the
interested reader. We note explicitly, however, that Question 1.3 remains open
even in the case κ = ω.

We anticipate that Question 1.3 has a negative answer. When seeking a
positive response, however, we have been drawn to the following question, which
is closely related to Theorem 2.8. Clearly, a positive answer to 6.1 will respond
positively also to 1.3 (in the case that Z is a Hausdorff space).

Problem 6.1. Let α ≥ ω, XI be a product space, Z be a space, and Y be
a dense subspace of XI such that πJ [Y ] is C(Z)-embedded in XJ for every
nonempty J ∈ [I]<α. If f ∈ C(Y,Z) depends on some nonempty J′ ∈ [I]<α,
must there be a nonempty J ∈ [I]<α such that f depends on J and fJ is
continuous?

Returning finally to Question 1.3, we remark that if a counterexample exists
then there will be a product space XI , a cardinal number α ≥ ω, a dense
subspace Y of XI , a space Z, and a function f ∈ C(Y,Z) such that πJ [Y ]
is C(Z)-embedded in XJ for every J ∈ [I]

<α, f depends on some nonempty
J ∈ [I]<α, and f does not extend continuously over XI . In particular, then,
fJ must be discontinuous for each J ∈ [I]

<α on which f depends.



On the continuity of factorizations 279

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Received October 2007

Accepted May 2008



280 W. W. Comfort, I. S. Gotchev and L. Recoder-Núñez

W. W. Comfort (wcomfort@wesleyan.edu)
Department of Mathematics and Computer Science, Wesleyan University, Mid-
dletown, CT 06459, USA

Ivan S. Gotchev (gotchevi@ccsu.edu)
Department of Mathematical Sciences, Central Connecticut State University,
1615 Stanley Street, New Britain, CT 06050, USA

Luis Recoder-Núñez (recoderl@ccsu.edu)
Department of Mathematical Sciences, Central Connecticut State University,
1615 Stanley Street, New Britain, CT 06050, USA