LH has nonempty products

Peter Selinger

March 15, 1994

1 The construction of binary products

We will show that LH, the category of topological spaces with local homeomorphisms, has binary products (in fact, it has arbitrary nonempty limits; see 2.3 below). This has been previously believed to be false, see e.g. [1, p. 50]. In the first definition we fix some notation:

Definition For an element x of a topological space X, we will write U_{x} for the set of all open neigborhoods of x. f : X → Y is a local homeomorphism if for every x ∈ X there exists a U ∈ U_{x} such that f|_{U} is a homeomorphism onto an open subset of Y. Denote by int A the interior of A ⊆ X. For any pair of maps ϕ : U → Y, ϕ′ : V → Y defined on open subsets U, V ⊆ X, define the (set) equalizer

eq(ϕ, ϕ′) := {x ∈ U ∩ V | ϕ(x) = ϕ′(x)}.

In the following, we will only consider open equalizers int eq(ϕ, ϕ′).

We will now construct the categorical product B of two spaces X, Y ∈ LH. The basic idea is to consider triples ⟨U, V, ϕ⟩ consisting of two open sets U ⊆ X, V ⊆ Y together with a homeomorphism ϕ : U → V. These will be the prototypes for open sets in B. The elements of B will arise as equivalence classes of quintuples (x, y, U, V, ϕ) with U, V and ϕ as before and x ∈ U, y ∈ V such that ϕ(x) = y. Let therefore

A := {(x, y, U, V, ϕ) | U ⊆ X open

V ⊆ Y open

ϕ : U → V a homeomorphism x ∈ U, y ∈ V, ϕ(x) = y }.

Strictly speaking, the tuple (x, y, U, V, ϕ) is completely determined by x, U and ϕ, but it is more convenient for us to write all the components. On A define (x, y, U, V, ϕ) ∼ (x, y, U′, V′, ϕ′) if there exists Ũ ∈ U_{x} with Ũ ⊆ U ∩ U′ and

ϕ|Ũ = ϕ′|Ũ.

Note that in this case, Ũ ⊆ int eq(ϕ, ϕ′), hence without loss of generality Ũ = int eq(ϕ, ϕ′), or

(x, y, U, V, ϕ) ∼ (x, y, U′, V′, ϕ′)  iff  x ∈ int eq(ϕ, ϕ′).

∼ is an equivalence relation on A: reflexivity and symmetry are obvious; for transitivity notice that if ϕ|Ũ = ϕ′|Ũ and ϕ′|Ũ′ = ϕ′′|Ũ′, then ϕ|_{Ũ∩Ũ′} = ϕ′′|_{Ũ∩Ũ′}. Define B = A/∼ and write [x, y, U, V, ϕ] for the ∼-class of (x, y, U, V, ϕ).

Now consider a topology on B: for any open sets O ⊆ X, P ⊆ Y and any homeomorphism ψ : O → P, define

⟨O, P, ψ⟩ := {[x, y, O, P, ψ] | x ∈ O, y ∈ P, ψ(x) = y}.

Note that ⟨O, P, ψ⟩ ⊆ B and that [x, y, U, V, ϕ] ∈ ⟨O, P, ψ⟩ iff x ∈ O and ϕ|Ũ = ψ|Ũ for some Ũ ∈ U_{x}. Let the topology on B be generated by the sets ⟨O, P, ψ⟩.

Lemma 1 The intersection of ⟨O, P, ψ⟩ and ⟨O′, P′, ψ′⟩ is again of the form ⟨O′′, P′′, ψ′′⟩, with

O′′ = int eq(ψ, ψ′)
ψ′′ = ψ|_{O′′} = ψ′|_{O′′}
P′′ = ψ′′(O′′).

In particular, the sets ⟨O, P, ψ⟩ form a basis, and not just a subbasis, of the topology on B, i.e. every open set is a union of basis sets. Proof: [x, y, U, V, ϕ] ∈ ⟨O, P, ψ⟩ ∩ ⟨O′, P′, ψ′⟩ iff x ∈ O ∩ O′ and ϕ|Ũ = ψ|Ũ and ϕ|Ũ′ = ψ′|Ũ′ for some Ũ, Ũ′ ∈ U_{x} iff x ∈ O ∩ O′ and ϕ|Ũ′′ = ψ|Ũ′′ = ψ′|Ũ′′ for some Ũ′′ ∈ U_{x}, or equivalently x ∈ Ũ′′ ⊆ int eq(ψ, ψ′) =: O′′ and ϕ|Ũ′′ = ψ′′|Ũ′′. This finally is the case if and only if [x, y, U, V, ϕ] ∈

⟨O′′, P′′, ψ′′⟩. □

We now consider the projection maps

π_{1} : B → X : [x, y, U, V, ϕ] ↦ x
π_{2} : B → Y : [x, y, U, V, ϕ] ↦ y.

Lemma 2 π_{1} and π_{2} are local homeomorphisms. More specifically π_{1} and π_{2} map ⟨U, V, ϕ⟩ homeomorphically onto U and V, respectively. Proof: We consider only π_{1}; the proof for π_{2} is entirely symmetric. Take an arbitrary point b = [x, y, U, V, ϕ] ∈ B. Then ⟨U, V, ϕ⟩ is a neighborhood of b and π_{1}|_{⟨U,V,ϕ⟩} is a one-to-one map onto U, with inverse

(π_{1}|_{⟨U,V,ϕ⟩})^{−1} : U → ⟨U, V, ϕ⟩ : x ↦ [x, ϕ(x), U, V, ϕ].

π_{1}|_{⟨U,V,ϕ⟩} is continuous: for open U′ ⊆ U we let V′ = ϕ(U′) and have

π_{1}|_{⟨U,V,ϕ⟩}^{−1}(U′) = {[x, y, U, V, ϕ] | x ∈ U′}
= {[x, y, U′, V′, ϕ|_{U′}] | x ∈ U′}
= ⟨U′, V′, ϕ|_{U′}⟩ ,

which is open.

π_{1} (and hence π_{1}|_{⟨U,V,ϕ⟩}) is an open map: by Lemma 1 it suffices to show this for basic open sets, and in fact we have π_{1}(⟨O, P, ψ⟩) = O. □

Lemma 2 establishes the existence of the diagram

\begin{matrix}B & [figure] & π_{1} & [figure] & [figure] & π_{2} & [figure] \\ X & & & Y. \end{matrix}

We will now show that this is in fact a product diagram. Given local homeomorphisms

[figure: Z; ρ_{1}; ρ_{2}] [figure: B; π_{1}; π_{2}] X  Y,

we construct f : Z → B as follows: for any z ∈ Z consider an open neighborhood W ∈ U_{z} on which both ρ_{1} and ρ_{2} are homeomorphisms.^{1} Let

x = ρ_{1}(z)
y = ρ_{2}(z)
U = ρ_{1}(W)  (1)
V = ρ_{2}(W)
ϕ = ρ_{2} ∘ (ρ_{1}|_{W})^{−1} : U → V.

Then [x, y, U, V, ϕ] ∈ B. Moreover, [x, y, U, V, ϕ] is independent of the particular choice of W, for if we choose, say, W′ ⊆ W and we construct [x′, y′, U′, V′, ϕ′] from W′, then x = x′, y = y′, U′ ⊆ U, V′ ⊆ V and ϕ′ = ϕ|_{W′} follow directly from the definition. Hence

f(z) = [x, y, U, V, ϕ]

is well-defined in this way for any z ∈ Z. We clearly have ρ_{1} = π_{1} ∘ f, ρ_{2} = π_{2} ∘ f.

Lemma 3 f : Z → B is a local homeomorphism. Proof: Starting with a z ∈ Z, choose a neigborhood W ∈ U_{z} on which f is a homeomorphism. Then the image f(W) is contained in ⟨U, V, ϕ⟩, with U = ρ_{1}(W), V = ρ_{2}(W), ϕ = ρ_{2} ∘ (ρ_{1}|_{W})^{−1} as before.

To conclude that f is a homeomorphism, we recall that W is by its choice homeomorphic to U via ρ_{1}, while U is the homeomorphic image of ⟨U, V, ϕ⟩ under π_{1} (by Lemma 2. Hence f|_{W} = (π_{1}|_{⟨U,V,ϕ⟩})^{−1} ∘ ρ_{1}|_{W} is a homeomorphism as well. □

Lemma 4 f is the unique morphism with

[figure: Z; ρ_{1}; ρ_{2}; f] [figure: B; π_{1}; π_{2}] X  Y.

Proof: Suppose g is another map making this diagram commute. Let z ∈ Z be arbitrary and pick W ∈ U_{z} such that ρ_{1}, ρ_{2}, f, g are all homeomorphisms on W. Moreover, pick W small enough such that g(W) is a basis open set in B, which can be done by the Lemma 1. Say g(W) = ⟨O, P, ψ⟩. Then g(z) = [x′, y′, O, P, ψ], for some x′ ∈ O, y′ ∈ P, while f(z) = [x, y, U, V, ϕ] and f(W) = ⟨U, V, ϕ⟩, where x, y, U, V, ϕ are as in (1). By commutativity and Lemma 2 we get

1This is where the construction does not generalize to the case of infinite products. However, infinite products do in fact exist; see 2.3 below.

x = π_{1}(f(z)) = ρ_{1}(z) = π_{1}(g(z)) = x′
y = π_{2}(f(z)) = ρ_{2}(z) = π_{2}(g(z)) = y′
U = π_{1}(f(W)) = ρ_{1}(W) = π_{1}(g(W)) = O
V = π_{2}(f(W)) = ρ_{2}(W) = π_{2}(g(W)) = P.  (2)

Finally, for any w ∈ W we have g(w) = [ρ_{1}(w), ρ_{2}(w), O, P, ψ] and f(w) = [ρ_{1}(w), ρ_{2}(w), U, V, ϕ], establishing

ψ(ρ_{1}(w)) = ρ_{2}(w) = ϕ(ρ_{1}(w)).

But since ρ_{1}|_{W} is onto O, this yields ψ = ϕ. Together with (2) this yields the uniqueness of f:

g(z) = [x′, y′, O, P, ψ] = [x, y, U, V, ϕ] = f(z).
□

So we have established

Theorem 5 LH has binary products. □

2 Remarks

2.1 Subcategories of LH

Since the maps π_{1}, π_{2} are local homeomorphisms, the product X × Y inherits from X and, independently, from Y, all local properties (i.e. properties that are reflected by local homeomorphisms). Among these are e.g. the separation properties T_{0} and T_{1}, local compactness, local metrizability etc. (notice, however, that being Hausdorff is not a local property). Hence the respective full subcategories of LH have products as well.

We get for example a nice result for posets by specializing to the full subcategory of LH consisting of T_{0} spaces in which arbitrary intersections of opens are open (this is a local property, too). This category is equivalent [1, 1.372] to the category POS^{l.i.} of posets with local isomorphisms (f : P → Q is a local isomorphism of posets if for any a ∈ P, f|_{↓a} : ↓a ⟶^{∼} ↓b). In this category, the product

of two posets P and Q has elements [a, b, f] where a ∈ P, b ∈ Q and f : ↓a ⟶^{∼} ↓b, ordered by [a, b, f] ≤ [a′, b′, f′] iff a ≤ a′, b ≤ b′ and f = f′|_{↓a}.

2.2 Almost a regular category (and more!)

Since LH also has pullbacks, equalizers and images and pullbacks transfer covers, it has all the properties of a regular category [1, 1.5]—with one important exception: it doesn’t possess a terminator. Despite the lack of a terminator, certain constuctions can be carried out as in a regular category, e.g. composition of relations is well-defined and associative, and we have a notion of a graph of a function as being an entire and simple relation [1, 1.564].

The latter comes out rather naturally in LH: the graph of f : X → Y is given by the subobject of X × Y

graph f = {[x, y, U, V, f|_{U}] ∈ X × Y }.

Conversely, a subobject G ⊆ X × Y is a graph iff all ⟨U, V, ϕ⟩ , ⟨U′, V′, ϕ′⟩ ∈ G satisfy ϕ|_{U∩U′} = ϕ′|_{U∩U′} (i.e. π_{1} simple), and X = ⋃_{⟨U,V,ϕ⟩∈G} U (i.e. π_{1} entire). The function f can then be recovered by ‘pasting’ all the local ϕ’s together.

2.3 LH is “conditionally complete”

As David Feldman has pointed out, the product of X and Y, when regarded as a sheaf over X, is given by the functor that associates to an open set U of X all local homeomorphisms from U to Y. While the description of binary products in section 1 avove does not immediately generalize to infinite products (see the footnote before Lemma 3), the sheaf description does. But unlike in the finite case, in the general (infinite) case there is no obvious no symmetric description of the product. One of the factors has to be preferred as the base space of the constructed sheaf. This asymmetry seems to be the deeper reason for the absence of an empty product, i.e. a terminator.

The presence of equalizers implies that every nonempty diagram in LH has a limit. Peter Freyd proposes to call this property “conditional completeness”.

One might hope that LH is also “conditionally exponential”, i.e. that certain exponential objects exist (clearly B^{0} doesn’t exist for any B, for it would be a terminator). However, upon closer inspection it turns out that there aren’t any exponentials in LH: given objects A, B, suppose there exists B^{A}. Now choose a nonempty C which has no open neighborhood homeomorphic to an open neighborhood in either A or B^{A} (e.g. choose C of cardinality bigger than that of A and B^{A} with the indiscrete topology). Then both (C, B^{A}) and C × A are empty and (C × A, B) is a singleton.

2.4 The right adjoint of the pullback functor

Instead of exponentials, we get a right adjoint for pullbacks in LH. This is generally true for categories whose slices are cartesian closed:

For any category A with pullbacks, and any f : A → B, consider the functor f^{#} : A/B → A/A defined by pullbacks:

[figure: D; f^{#}x; A; f; C; x; B]

Proposition 6 A/B is cartesian closed iff f^{#} has a right adjoint for all f : A → B. Proof: This is a consequence of 1.854 in [1] and the fact that A/A ≃ (A/B)/(f). □

Since LH/Y = Sh(Y) is cartesian closed, we know that in LH the pullback functor has a right adjoint. Will it help us if we understand what this adjoint is? … to be continued …

References

[1] P. J. Freyd and A. Scedrov. Categories, Allegories, volume 39 of North-Holland Mathematical Library. North-Holland, 1989.