PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 124, Number 4, April 1996

RIGHT ADJOINT FOR THE SMASH PRODUCT FUNCTOR

FRANCESCA CAGLIARI

(Communicated by Andreas R. Blass)

Abstract. The smash-product functor (–) (X, x0) in the category Top of ∧ pointed topological spaces has a right adjoint for any choice of the base point∗ x0, if and only if the topological space X is quasi-locally compact, that is, if and only if the product functor (–) X has a right adjoint in the category Top of topological spaces. ×

Introduction AspaceXis cartesian in the category of topological spaces and continuous maps if the product functor (–) X has a right adjoint. This means that there exists a proper and admissible topology× on the space of maps Y X between X and Y (for any topological space Y ) [D]. Cartesian objects in Top were characterized by Day and Kelly [D-K]. They are the quasi-locally compact spaces [H-L]. The problem of the existence of a proper and admissible topology on the function (X,x0) X space (Y,y0) consisting of the maps of Y preserving base points is related to the adjointness of the smash-product functor. It is known that this functor has a right adjoint whenever X is locally compact and Hausdorﬀ; in this case the (X,x0) topology on (Y,y0) is the compact open topology [M]. In this paper, it is proved that the spaces (X, x0) for which the functor (–) ∧ (X, x0) has a right adjoint are exactly the spaces X which are cartesian in Top, independently of the choice of x0. That is, the existence of a proper and admissible (X,x0) topology on (Y,y0) for any (Y,y0) is equivalent to the existence of a proper and admissible topology on the whole space of maps from X to Y , for any Y .

Smash-product and adjunction

We can consider, in Top , the endofunctor (–) (X, x0) and ask when it has a right adjoint. When it exists∗ we will call it ∧. G(X,x0) In the case of X cartesian in Top, we indicate by Y X the power object and by (X,x0) X X (Y,y0) the subspace of Y given by f Y f(x0)=y0 with base point { ∈ | } the constant y0-valued map.

Theorem 1. If X is cartesian in Top,then(–) (X, x0) has a right adjoint, for ∧ (X,x0) every x0 in X. Moreover (Y,y0)=(Y,y0) . G(X,x0)

Received by the editors November 3, 1993 and, in revised form, August 19, 1994. 1991 Mathematics Subject Classiﬁcation. Primary 54B30; Secondary 18A40.

c 1996 American Mathematical Society

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Proof. Suppose X is cartesian in Top and take any space Y ; lete ˆ: Y X X X × → Y be the evaluation map. Consider in Y the subspace (X,x0)(Y,y0)= f X G { ∈ Y f(x0)=y0 and the restriction e1 ofe ˆ to (Y,y0) (X, x0)whichisa | } G(X,x0) × map in Top .Themape1is compatible with the quotient in the deﬁnition of the ∗ smash-product, and so we can consider the map e: (Y,y0) (X, x0) (Y,y0) G(X,x0) ∧ → induced by e1.Wenotethate(f,x)=f(x). Let f :(Z, z0) (X, x0) (Y,y0)beamapinTop and consider the quotient ∧ → ∗ p:(Z, z0) (X, x0) (Z, z0) (X, x0) which gives the smash-product. Since X is × → ∧ cartesian in Top, related to fp:(Z, z0) (X, x0) (Y,y0)thereisan(fp)1: Z X × → → Y such thate ˆ((fp)1 idX )=fp.Themap(fp)1 preserves the base points and its × image is a subspace of (X,x0)(Y,y0), so we can factor (fp)1 through the inclusion XG of (X,x0)(Y,y0)inY and consider the ﬁrst factor (fp)2 as a map in Top .In G ∗ such a way we obtain (fp)2 idX :(Z, z0) (X, x0) (Y,y0). By construction × × → (fp)2 idX is compatible with the quotient p, and the proof is complete. ×

Theorem 2. If the functor (–) (X, x0) has a right adjoint, then (X,x0)(Y,y0) is a ∧ G (X,x0) space whose underlying set is in natural bijective correspondence with (Y,y0) ,

the counit of the adjunction is the map e: (X,x0)(Y,y0) (X, x0) (Y,y0) such that e(f,x)=f(x)and the base point correspondsG to the∧ constant function→ valued at y0.

Proof. Let D2 be the space with two points a, b and the discrete topology. By the (D2,a) (X,x0) (D ,a) adjunction, there is a bijection between (Y,y0) ∧ and ( (Y,y0)) 2 , G(X,x0) andontheotherside(D2,a) (X, x0) is homeomorphic to (X, x0)and (D2 ,a) ∧ (G(X,x0)(Y,y0)) is in bijection with (X,x0)(Y,y0); so the ﬁrst part of the theorem is proved. G Any map f :(X, x0) (Y,y0) can be considered as a map from (D2,a) → ∧ (X, x0)into(Y,y0). As a consequence, by the adjunction, for any f,thereisan f1:(D2,a) (Y,y0) such that e(f1 idX )=f;soe(f,x)=f(x). Finally, →G(X,x0) ∧ given the one point space ,andthemaph: (X, x0) (Y,y0), there is a • •∧ → map h1 : X,x0)(Y,y0) such that h1( ) is the base point of (X,x0)(Y,y0). This completes•→G the proof. • G We denote by S the Sierpinski space with the two points 0 and 1 and 0 the { } nontrivial open set. If the functor (–) (X, x0) has a right adjoint, as a consequence of Theorem 2, (S, 0) can be identiﬁed∧ with the set of the open sets U of X G(X,x0) such that x0 U and base point the open set X. On the other hand, (S, 1) ∈ G(X,x0) can be identiﬁed with the set of the open sets U of X such that x0 / U and base point the empty set. ∈ The following Lemma characterizes convergent nets of the spaces (S, 0) G(X,x0) (respectively, (X,x0)(S, 1)), while Lemma 4 proves that the open sets of these spaces are Scott-openG [H-L].

Lemma 3. Suppose (–) (X, x0) admits a right adjoint. A net Ui converges to U in (S, 0) (respectively,∧ (S, 1)) if and only if : G(X,x0) G(X,x0) ( ) for each x U and for each net xλ converging to x in ∗ ∈ X, there is an i0 and a λ0 such that xλ Ui, for every ∈ i>i0 and λ>λ0. Proof. Let Ui converge to U in (S, 0) (respectively, (S, 1)),x U and G(X,x0) G(X,x0) ∈ xλ converge to x in X. Consider the counit of the adjunction e: (S, 0) G(X,x0) ∧ License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use RIGHT ADJOINT FOR THE SMASH PRODUCT FUNCTOR 1267

(X, x0) (S, 0) (respectively, (S, 1)) and the quotient map p: (S, 0) → G(X,x0) × (X, x0) (S, 0) (X, x0) (respectively, (S, 1)). Since (Ui,xλ)converges →G(X,x0) ∧ to (U, x)andthemapep is continuous with ep(U, x)=0,thenep(Ui,xλ)converges to 0. In S the nets converging to 0 are eventually constant, so there is an i0 and a λ0 such that, for every i>i0 and λ>λ0,ep(Ui,xλ) = 0, that is xλ Ui. ∈ Vice versa, suppose Ui is a net in (X,x0)(S, 0) and U (X,x0)(S, 0) (respectively, (S, 1)) fulﬁlling condition ( G). Consider the space∈GI ,whereIis the G(X,x0) ∗ ∪ {•} direct set of the net Ui, is a maximum point whose base-neighborhoods are the • sets of the form Ij = i I i j ,j I,andthepointsofIare isolated. {•} ∪ { ∈ | ≥ } ∈ We can assume, without changing the nature of the net Ui, that there exists a point i0 I, such that Ui = X (respectively, Ui = ). Moreover, we consider the map ∈ 0 0 ∅ α:(I ,i0) (X,x0)(S, 0) (respectively, (X,x0)(S, 1)) so deﬁned by α(i)=Ui, α( )=∪ {•}U. We prove→G that the map α is continuous,G which implies the convergence • of the net Ui to U. By the existence of the right adjoint, α is continuous if and only if the corresponding map e(α IdX )=α:(I ,i0) (X, x0) (S, 0) (respectively, (S, 1)) is continuous; so∧ we will prove the∪ continuity {•} ∧ of α.Tothisaim→ we can consider the quotient map q :(I ,i0) (X, x0) (I ,i0) (X, x0) induced by the smash-product and prove∪ {•} the continuity× of→αq.∪ By {•} the adjunction∧ 1 (αq)− (0) = (i, x) x Ui,i I ( ,x) x U . Since every point of { | ∈ ∈ }∪{• | ∈ } I is isolated and each Ui is open, the set (i, x) x Ui,i I is open since { | ∈1 ∈ } it is union of open sets, therefore each (i, x)of(αq)− (0) belongs to its interior. 1 Take now ( ,x) (αq)− (0) (that is x U); the topology deﬁned on (I ) and condition• ( )∈ implies that any net converging∈ to ( ,x) in the product∪ space {•} ∗ 1 • (I ,i0) (X, x0) is eventually in (¯αq)− (0), that is, ( ,x) belongs to the interior ∪{•} 1 × 1 • of (αq)− (0). We can conclude that (¯αq)− (0) is open and so, α is continuous.

Lemma 4. Suppose (–) (X, x0) admits a right adjoint, and suppose H is open in (S, 0) (respectively,∧ (S, 1)); then H is Scott-open [H-L], i.e.: G(X,x0) G(X,x0) (a) If U, U 0 are open in X, U H and U 0 U (with x0 / U 0 when H is open in ∈ ⊇ ∈ (S, 1)),thenU0 H. G(X,x0) ∈ (b) If V = Ui i I is a family of open subsets of X and U = Ui i I H, there exists{ | a∈ ﬁnite} subfamily of V whose union belongs to H. { | ∈ }∈ S Proof. (a) Suppose U, U 0 open in X, U H and U 0 U (x0 / U 0 in the case when ∈ ⊇ ∈ H is open in (S, 1)). The constant sequence whose value is U 0 converges to G(X,x0) U in (X,x0)(S, 0) (respectively, (X,x0)(S, 1)) because it fulﬁls the condition ( )in LemmaG 3. Since H is open andGU belongs to H, there must be an element of∗ the constant sequence belonging to H; consequently U 0 H. ∈ (b) Let U = Ui i I H.IfHis open in (S, 0), it follows that { | ∈ }∈ G(X,x0) x0 U; therefore there exists an i0 such that x0 Ui .Considerin (X,x )(S, 0) ∈ S ∈ 0 G 0 the net whose direct set is (i1,i2,...,in) i1,i2,...,in I with the relation { | ∈ } (i1,i2,...,in) (j1,j2,...,jm)ifUi Ui Ui Uj Uj Uj and 1 ∪ 2 ∪···∪ n ⊇ 1 ∪ 2 ∪···∪ m image of (i1,i2,...,in)equaltoUi0 Ui1 Ui2 Uin. This net, according to ( ), converges to U in (S, 0); as∪ a consequence,∪ ∪···∪ since H is an open set which ∗ G(X,x0) contains the limit of the net, there exists (i1,i2,...,in) such that Ui Ui Ui 0 ∪ 1 ∪ 2 ∪···∪ Uin H.WhenHis open in (X,x0)(S, 1),x0 / U and therefore, x0 / Ui,foreach i I∈. In this case, we can considerG the net with∈ the same direct set∈ as above and ∈ with the image of (i1,i2,...,in)equaltoUi Ui Ui . The same argument of 1∪ 2∪···∪ n the ﬁrst case proves that there is (i1,i2,...,in) such that Ui Ui Ui H. 1 ∪ 2 ∪···∪ n ∈

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Deﬁnition 5. AspaceXis quasi-locally compact if for every x in X and for every neighbourhood U of x there is a neighbourhood of V U of x such that every open cover of U has a ﬁnite subcover of V . ⊆ Theorem 6. Let X be a topological space. The following statements are equivalent : (1) There exists x0 in X such that the functor (–) (X, x0): Top Top has a right adjoint. ∧ ∗ → ∗ (2) X is cartesian in Top, that is X is quasi-locally compact. (3) For any x0 in X,thefunctor(–) (X, x0): Top Top has a right adjoint. ∧ ∗ → ∗ Proof. (1) (2) Since the continuity of the counit of the adjunction e: (S, 0) → G(X,x0) (X, x0) (S, 0) implies the continuity of the evaluation map e0 : (X,x0)(S, 0) ∧ → 1 G × (X, x0) (S, 0), it follows that (e0)− (0) is an open set with e0(W, x0)=0,forany → W (X,x0)(S, 0). Take∈G a point x X and ﬁx U open in X with x U. ∈ ∈ Suppose x cl x0 ;thenx0 U. Consequently U belongs to (X,x0)(S, 0) ∈ { } 1 ∈ G and e0(U, x) = 0. Since (e0)− (0) is open, there is an H,openin (X,x0)(S, 0), with 1 G U H and a neighbourhood V of x such that (e0)− (0) H V .Sincee0(W, y)=0 when∈ y W , any element of H contains V . Now, consider⊇ × an open cover of U , ∈ Ui i I and consider U 0 = Ui i I .SinceU0 U, by Lemma 4a), U 0 H { | ∈ } { | ∈ } ⊇ ∈ and by Lemma 4b) there are i1,i2,...,in I such that Ui Ui Ui H S ∈ 1 ∪ 2 ∪···∪ n ∈ andthisimpliesthatUi Ui Ui V. 1 ∪ 2 ∪···∪ n ⊇ Suppose now x/cl x0 . Therefore, there is an open A of X, such that x A and ∈ { } ∈ x0 / A. The open set U A (S, 1); the map e: (S, 1) (X, x0) ∈ ∩ ∈G(X,x0) G(X,x0) ∧ → (S, 1) is continuous and e(U A, x) = 0. Replacing U 0 = Ui i I by ∩ { | ∈ } U 0 = Ui A i I in the argument used before proves that there exists a neighbourhood{ ∩ V| of∈x with} U A V , such that any open coverS of U (and then of U SA), admits a ﬁnite subcover∩ ⊇ for V . (2)∩ (3) Theorem 1. (3) → (1) Trivial. → Acknowledgment The author wishes to thank the referee, for his helpful suggestion and for this patience. References

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