Proc. Camb. Phil. Soc. (1970), 67, 553
POPS 67-60
Printed in Great Britain
,
553
On topological quotient maps preserved by pullbacks or products
B. J. DAY AND G. M. KELLY
University of New South Wales, Australia
(Received 3 February 1969)
1. Introduction
We are concerned with the category of topological spaces and continuous maps.
A surjection/: X-+ Y in this category is called a quotient map if G is open in Y whenever f~xG is open in X. Our purpose is to answer the following three questions:
Question 1. For which continuous surjections / : X - > Y is every pullbackf of / a
quotient map ?
Question 2. For which continuous surjections f: X^> Y is f x lz: X x Z^* Y x Z &
quotient map for every topological space Z ? (These include all those / answering to
Question 1, since/x lz is the pullback of/by the projection map Y xZ^- Y.)
Question 3. For which topological spaces Z is f x lz: X x Z-+ Y x Z a, quotient map
for every quotient map /?
Had we posed the corresponding questions for the category of Hausdorff spaces
instead of the category of topological spaces, known results would have sufficed to
answer them. For a topological space Z to answer to Question 3, it is sufficient that
each point of Z have a fundamental system of compact neighbourhoods (Brown (1),
4.3.2); if Z is Hausdorff, its local compactness is also necessary, even if we only consider those / with X and Y Hausdorff (Michael (4)). For a continuous surjection
/ : X -¥• Y to answer to Question 2, it is sufficient that it satisfy the following condition
[A'), which is moreover necessary when Y is Hausdorff, even if we only consider
Hausdorff Z (HajekU), with correction (3)):
(A') Given any filter 0 on Y and any yeY adherent to 0, there is some x ef~hj with x
adherent tof^1®.
This result was rediscovered by Michael (5), who also proved the equivalence of (A1)
with the following condition (A):%
(A) Given yeY
and an open covering (Ga)aeA
of f~xy, there is a finite set
{av ..., ocn} <= A such that fGai u ... U fGXn is a neighbourhood of y.
Finally, although Michael does not explicitly consider pullbacks, it may be deduced
from Theorem 1-2 and Proposition 3-1 of (5) that a n / satisfying (A) answers not only
to Question 2 but also to Question 1.
None of these known results, however, quite suffices to answer the questions as
we have posed them for the topological category; we give the answers in Theorems
1-3 below.
f The pullback of f :X ->• Y by v.Z -> Y is g :W -+ Z, where W is the subspace of Z x X
consisting of those (z, x) with vz = fx, and g is the projection.
X Maps/ satisfying (A) were called bi-quotient maps in (5).
36-2
554
B. J. DAY AND G. M. KELLY
2. Statement of results
THEOREM 1. Let f:X-> Y be a continuous surjection. Then every pullback of f is a
quotient map if and only if f satisfies condition (A).
If K is a subset of a topological space Y, let us write K* for the intersection of all
the open sets containing K. Note that K* = K for all K <= Y if and only if Y is a Tx
space.
THEOREM 2. Letf: X—-Y be a continuous surjection. Thenfx lz: XxZ->Y xZ is a
quotient map for all topological spaces Z if and only iffis a quotient map and satisfies the
following condition (B):
(-B) Given yeY, a neighbourhood V of y, and an open covering (6ra)aeA of f~xV, there
is a finite set {als ...,a n } c A such that {fGXi U ... VfGaJ* is a neighbourhood of y.
Now let Zbea, topological space and write Q for the set of open sets of Z. It is easy
to see that we get a topology on Q if we take the open sets to be exactly those H <= Q
which satisfy the following two conditions:
01. If V, F'efiwith V <= V and if FeH then V'eH.
02. If Va e Ofor aeA and if UaFae.ff, then there is afiniteset {a1; ...,an} <= A with
THEOREM 3. Let Z be a topological space and let Q, be as above. Then fxlz:
XxZ->YxZisa
quotient map for all quotient maps f:X-*-Y if and only if Z satisfies the following condition (G):
(C) Given W e Q, and z e W, there is an open set H in Q, with W eH, such that OVEHV
is a neighbourhood of z in Z.
3. Discussion of results
Note that in Theorem 2 we have to assert explicitly that/is a quotient map, for
this is not a consequence of (B) as it is of (A); indeed {B) is vacuous if Y has the trivial
topology.
If Y has the trivial topology and/:X->• Y is a quotient map,/automatically satisfies (B); but it need not satisfy (.4). As an example let X = Z x {0,1} where Z is the
set of integers, let Y = Z, and let /:Z x {0,1}->Z be the projection. Define an involution t: X^-X by t(n, 0) = (n+ 1,1), t(n, 1) = (n— 1, 0); and give X the topology
in which the open sets are those stable under t. Give Y the topology that makes / a
quotient map; this is clearly the trivial topology. The condition (A) is then false for the
covering of/^O by the single open set {(- 1,0), (0,1), (0, 0), (1,1)}.
Even when Y is a Tx space, so that K* — K for all K c Y, it is possible for a
quotient map / to satisfy (B) and not (A). Take f:X^-Y
to be the projection
Z x {0,1}-»Z as in the example above, but this time give X the topology generated
by the following sets Pn(neZ) and Q: the set Pn is (Z — {n}) x {0,1}, and the set Q is
(2Zx{0})U({0}x{0,l}),
where 2Z denotes the even integers in Z. Give Y the topology that makes/a quotient
map; clearly this is the 3\ topology in which the closed sets other than Y itself are
the finite sets. It is easy to see that (B) is satisfied, but (.4) is false for the covering of
/ - 1 0 by the single open set Q.
On topological quotient maps
555
On the other hand, since K* = K for K <= Y when Y is Hausdorff, Proposition 2-1
of (5) gives at once:
PROPOSITION 4. If Y is Hausdorff and the continuous surjection f:X^-Y
satisfies
(B) then f satisfies (A).
We now turn to Theorem 3 and discuss possible simplifications of condition (O).
In our terminology neither a compact space nor a regular space need be Hausdorff.
It is usual to call a space Z locally compact if each point of Z has a compact neighbourhood. Let us call Z properly locally compact if each point of Z has a fundamental
system of compact neighbourhoods. Clearly a locally compact Z is properly locally
compact if it is regular, and a fortiori if it is Hausdorff, since a locally compact Hausdorff space is regular.
5. (i) If Z is properly locally compact it satisfies condition (C).
(ii) If Z is regular and satisfies condition (G) it is properly locally compact.
(iii) / / Z is Hausdorff and satisfies condition (C) it is regular.
Of course (iii) above follows, given our Theorem 3, from Theorem 2-1 of (4), but we
give a direct proof. We do not know whether a non-regular space satisfying (C) need
be properly locally compact.
Proof of (i). Given We Q and zeW, we can find by hypothesis a compact neighbourhood K of z with KczW. Take H to be {Ve il\K <= V}, which clearly satisfies
0 1 and 02. Then WeH; and C\VeH^' which contains K, is a neighbourhood of z.
Thus (C) is satisfied.
Proof of (ii). Given WeQ, and zeW, let H be an open set in £2 as in condition (C).
Since Z is regular, the neighbourhood OVEH V of Z contains a closed neighbourhood
K of z; we assert that K is compact. If (Ga) is an open covering of K then (Ga) together
with W — K is an open covering of W. Since WeH it follows first from 01 that
and then from 02 that (W-K) u G«zxU ... [)GaneH for some
(W-K) U [JAeH
finite {<*!,..., ocn}. Since K by its definition is contained in every F e H, it follows finally
PROPOSITION
Proof of (iii). Given W e Q and zeW, let H be an open set in Q, as in condition (C),
and consider the neighbourhood N = f]VeH V of z. It suffices to prove that the closure
N of N is contained in W. Suppose then that yeZ—W. For each we W we can find,
because Z is Hausdorff, an open set Aw containing w whose closure does not contain
y. Since \JWAW => W and WeH, it follows from 0 1 and 02 that AWiU...[) AWneH
for some finite {wv ...,wn}. Therefore N c AWi u ... U AWa, whence yeZ — N, as
required.
4. Proofs
Proof of Theorem 1. (A) is necessary. Given y and (Ga) as in condition (A), set
Ha = (Y-fGa) U {y}. Let 0 be the filter on Y generated by all the Ha for a e A together with all the neighbourhoods of y in Y. Let Y' have the same underlying set as
Y, but with the following topology :Bc Y' is open if and only if either y $ B or B e 0.
i s / : X ' - > Y', where
Then 1: Y'-*• Y is continuous; the pullback by this of f:X^-Y
X' has the same underlying set as X and the coarsest topology for which 1: X' ->• X
and/: X' -> Y' are both continuous.
556
B . J . D A Y AND G. M.
KELLY
x
Ga is open in X' because it is open in X;f~ Ha is open in X' because Ha is open in
Y'; hence \Ja{Ga f]f-xHa) is open in X'. But \Ja{Ga nf+Ha) = \Ja(Ga nf~xy) =f~x y; it
follows that {y} is open in Y', since/: X' -»- Y' is a quotient map by hypothesis. Therefore {y} e 0 ; in other words, there are a finite set {av ..., an} c A and a neighbourhood
F of y in Y with {2/} = F n •H'o, D... fl #<*„• From the definition of Ha it follows that
/6r ai U ... UfGan contains F, and is therefore a neighbourhood of y.
(A) is sufficient. (We give a direct proof although, as we have pointed out above,
this can be deduced from the results of (5).) Consider a pullback diagram
u
Z
>Y
V
Note that W is the subspace of Z x X consisting of those (z, x) with vz = fx, and that
g, u are the projections.
Let A cz Z with g~xA open; we are to prove that A is open. Let then zeA; we are to
prove that A is a neighbourhood of z.
For each x ef~yvz we have (z, x) e g~xA; since g~xA is open in W there are open sets
Bx in Z containing z and Cx in X containing x such that (Bx x Cx) n W c g~xA; this
is to say that Bx n v~xfCx <= .4.
Since the G,. for a; ef~xvz form an open covering of f~ljvz, we can by (A) find
a^, ...,xnef^1vz such that /GXi U ... U/G^ is a neighbourhood of wz. It follows that
v~1fGXi U ... [)v~1fCXn is a neighbourhood of z, and therefore that
(Bx n . • • n Bx ) n {v-xfCx u . . . u v~xfCx)
is a neighbourhood of z; hence .4, which contains this last set, is also a neighbourhood
of z.
Proof of Theorem 2. (B) is necessary. Given y, F, and (Ga) as in condition (B), we may
without loss of generahty suppose F to be open (otherwise replace F by its interior)
and suppose Ga to be contained i n / - 1 F (otherwise replace Ga by Ga Of~xV).
Let Z be the set of open sets of Y and for each a e A set Ha = {WeZ\fGa <= W};
note that VeHa for each a. Let 0 be the filter on Z generated by all the i/ a for aeA,
and give Z a topology by taking Q c Z to be open if and only if either F ^ Q or Q e 0.
In XxZ consider the subset S = {(x,W)\fxeW); we assert that # is open. If
| F, then a; lies in the open set f~xW, W lies in the open set {W},
(x, W) eS with W ==
and f~xW x {W} <= #. If (x, F) e # we have xef~xY and so a; lies in the open set Ga
for some a; also F lies in the open set Hx, and GaxHa <= $.
$ is the inverse image under / x l z : J x Z ^ - 7 x Z o f the subset T = {(y, W)\t]e W)
of Y x Z. Since/x lz is a quotient map by hypothesis, it follows that T is open. Since
(y, V) e T, there are an open set P in Y containing y and a finite set {a x ,..., a n } c: A
such that P x (Ha n ... D !?« ) c J7. In other words, any open set W in Y which
contains fGa U...\jfGa also contains P . Therefore (fGa U ... UfGan)* is a neighbourhood of y.
On topological quotient maps
557
1
(B) is sufficient. Suppose that B <=• YxZ with C = (fx l ) " ^ open in X xZ; we
are to prove that B is open. Let then (y, z)eB; we are to prove that B is a neighbourhood of (y, z).
For each £eZ write C5 for {xeX\{x,QeC} and B^ for fye r|(^£)e-B}» s o that
Cj = /"^Bj. Then (7j is open because 0 is open, and hence B^ is open because / is a
quotient map.
For each x e Cz we have (x,z)eC and because C is open there are open sets Ox in X
containing x and Hx in Z containing z such that GxxHx<^ G. The sets Gx for a; e Cs
form an open covering of Gz = f~1Bz, and Bz is a neighbourhood of y; therefore by (JS)
we can find x1} ...,xneCz such that (fGXi U ... UfG^J* is a neighbourhood of y. We
complete the proof by showing that the neighbourhood
(fGXiu...ufGXn)*x(HXin...nHj
of (y, z) is contained in B.
If £eHXi n ... n #*„ then Gx. <= Cj for 1 ^ i ^ n, and so /<?x. <= /C f = 5^; therefore fGXi U ... U/G^ <= 5 f , and since B^ is open it follows that (fGXi U ... UfGxJ* <= JBj,
as required.
Proof of Theorem 3. We first establish some auxiliary propositions. We suppose
throughout that Z is a topological space and that £i is its set of open sets topologized
by taking the open sets of £1 to be those satisfying 0 1 and 02. We denote by T the
subset {(V,z)\ze V} of Q x Z. We also have occasion to consider topological spaces
with the same underlying set as Q but with possibly different topologies; these we
denote by modifications of Q, such as Q.', Qa. If Q.' is such a space we call its topology
admissible if T is open in Q' x Z. It is immediate that condition (C) may be written
equivalently as:
(C) The topology of Q. is admissible.
PROPOSITION 6. A function g: Y-+Q.from a topological space Y to Q. is continuous
if the set {(y, z)\zegy} is open in YxZ.
Proof. Let yeY and suppose that gyeH where H is open in Q.
For each zegy, because the set G = {(y,z)\zegy} is open, we can find an open set
Az in Y containing y and an open set Bz in Z containing z such that Az x Bz <= G.
Thus Bs <= gn for each TJEAZ.
Since yeAz we have in particular that Bs <=• gy for each zegy. Thus gy = \JssgyBs.
Since gyeH and H is open in Q it follows from 02 that we can find zx, ...,znegy with
Bz u...l)Bz eH. Now Bz u... UBZn<= grj for each tj in the neighbourhood
Az n ... n Az of y, so that for each such ij we have gtjeH by 0 1 . This proves the
continuity of g at y.
PROPOSITION 7. The topology of Q is the infimum of the set of admissible topologies.
Proof. If D.' has an admissible topology, the set T is by definition open in Q.' x Z,
so that the function 1: £2'->Q is continuous by Proposition 6. Thus the topology of
Q is coarser than any admissible topology.
Now let Q" have a topology coarser than any admissible topology; it remains to
show that the topology of Q" is coarser than that of Q.. Let then H be open in Q";
we are to show that H satisfies 01 and 02.
558
B. J. DAY AND G. M. KELLY
Suppose that [JaVaeH, where the Va are open in Z. Write W for \JaVa, set
Pa = {Ve Q\Va c V}, and let © be the filter on Q. generated by the Pa. Write Q' for Q
retopologized as follows: E <= Q.' is open in Q' if and only if either W$R or B 6 0.
The set T is open in Q.' x Z. For if z e V with F 4= W then T7 contains the neighbourhood {V} x V of (V,z) in. £1' x Z; while if ze W then zeF^ for some a and T contains
the neighbourhood Pa x Va of (W, z) in D' x Z. Thus the topology of Q.' is admissible
and therefore finer than that of Q".
It follows that H is open in Q,', since it is open in Q". Thus there is a finite set
{a1( ...,«„} with Pai n ... fl Pan <= H. This means that VeH whenever V is open and
V =>VXlU ...U Van; so clearly H satisfies both 0 1 and 02.
The condition (C) is necessary in Theorem 3. Give the set of open sets of Z all possible
admissible topologies, and denote the resulting topological spaces by Qa, where a
runs through some set A. Let SO a be the disjoint union of the Q a , with the sum topology
—that is, a set G is open in SO a if and only if each G n Oa is open in Q.a. Let/: SQ a -> O
be the function whose restriction to each Qa is 1. Then/is a quotient map by Proposition 7. By hypothesis, then, the m a p / x 1: (2Q a ) x Z-+Q, x Z is also a quotient map.
But ( S H J x Z = S(Q a x.Z), again with the sum topology. Since T is open in each
Q a x Z, it follows that T is open i n f l x Z ; this is condition (C), which is equivalent to
(C).
TAe condition (G) is sufficient in Theorem 3. L e t / : X - > F be a quotient map, let
B c 7 x 2 , and let C = (/x l ) - 1 ^ be open in J x Z . We are to prove that B is open.
For yeY write <??/ for the subset {zeZ\ (y, z) e B) of Z. Then
gfx = {zeZ|(/z,z)e£} = {ze.Z|(z, Z )eC},
which is open because C is open. Since/is surjective, gy is open for each y, so that g is a
function g: Y-> £2. The composed function gf:X->Q, is continuous by Proposition 6,
since C is open. Since / is a quotient map, g:Y->Q. is continuous; whence
gx 1: Fxi5->-Qx.Zis also continuous. By condition (C),in its equivalent form (C),
T is open i n Q x Z ; hence (gr x I)- 1 ? 7 is open in 7 x Z. But (gr x I)- 1 ! 1 = 5 .
REFERENCES
(1) BROWN, R. Elements of modern topology. London (1968).
(2) HAJBK, O. Notes on quotient maps. Comment. Math. Univ. Carolinae 7 (1966), 319-323.
(3) HAJEK, O. Correction to notes on quotient maps. Comment. Math. Univ. Carolinae 8 (1967),
171.
(4) MICHAEL, E. Local compactness and cartesian products of quotient maps and ^-spaces.
Quart. J. Math. Oxford Ser., to appear.
(5) MICHAEL, E. Bi-quotient maps and cartesian products of quotient maps. Ann. Inst. Henri
Poincare, Sect. B, to appear.