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The geometric realization of a simplicial Hausdorff space is Hausdorff

It is shown that the thin geometric realization of a simplicial Hausdorff space is Hausdorff. This proves a famous claim by Graeme Segal that the thin geometric realisation of a simplicial k-space is a k-space.


The main problem
In one of his many landmark papers [3], Graeme Segal introduced the geometric realization functor for simplicial spaces, which he called the "thin" realization functor in the subsequent article [2]. He claimed that the thin geometric realization of a simplicial space which is compactly-generated Hausdorff degreewise must be compactly-generated Hausdorff. However, while it is essentially obvious that the geometric realisation of a simplicial compactly-generated space must be compactly-generated, the Hausdorff property is a whole different matter since cocartesian squares are implicit in the definition of the thin geometric realization and they are known to behave badly with respect to separation axioms. At the time of [3], Segal's claim was thought to be dubious and no convicing proof of it ever appeared in the litterature. This difficulty brought some to turn away from k-spaces and work with weak-Hausdorff compactly-generated spaces instead. It is much easier indeed to show that the geometric realization of a compactly-generated weak-Hausdorff space is itself compactly-generated weak Hausdorff (this was done by Gaunce Lewis in the appendix of his PhD thesis). In the following pages, we will prove that Segal was right after all!

Definitions and notation
In this paper, we will use the French notation for the sets of integers: N will denote the set of natural numbers (i.e. non-negative integers), and N * the one of positive integers. Recall the simplicial category ∆ whose objects are the ordered sets [n] = {0, 1, . . . , n} for n ∈ N and whose morphisms are the non-decreasing maps, with the obvious compositions and identities. All the morphisms are composites of morphisms of two types, namely the face morphisms δ k i : for k ∈ N and i ∈ [k + 1], and the degeneracy morphisms for k ∈ N * and i ∈ [k − 1]. See [1] for a comprehensive account.
There is (covariant) functor ∆ * : ∆ → Top which sends [n] to the n-simplex A simplicial space is a contravariant functor ∆ → Top. Given such a functor, we set A n := A([n]) for any n ∈ N. For k ∈ N, we will write d k i := A(δ k i ) for i ∈ [k+1] (the face maps of A), and s k i := A(σ k i ) for i ∈ [k−1] (the degeneracy maps of A). When no confusion is possible, we will simply write δ i instead of δ k i , σ i instead of σ k i , d i instead of d k i and s i instead of s k i . If δ is a morphism in ∆, we will also write δ * instead of A(δ).
For n ∈ N, a point x ∈ A n is said to be degenerate when in the image of some s i . For every n ∈ N, we thus have a natural map π n : A n × ∆ n → |A|.
Our simple aim here is to prove the following theorem: Theorem 1.1. Let A be a simplicial space and assume that A n is Hausdorff for each n ∈ N. Then |A| is Hausdorff.
The proof, although very technical, has a very straightforward basic strategy: we will give a general construction of "flexible" open neighborhoods for the points of |A| (see Section 2 for the construction and Section 3 for the proof of openness), and then show that those neighborhoods may be used to separate points (Section 4). In the rest of the paper, A denotes an arbitrary simplicial space (no separation assumption will be made until Section 4).

Constructing open subsets in a geometric realization
In the whole section, we fix an integer n ∈ N, a non-degenerate simplex x ∈ A n and a point α ∈ ∆ n ∂∆ n . Our goal is to give a general construction of nontrivial open neighborhoods of π n (x, α) in |A|. This will be done by constructing, is an open subset of |A| which contains π n (x, α) and its inverse image by π k is V k for each k ∈ N.
Remark 1. Since every morphism in the simplicial category is a composite of face and degeneracy morphisms, a family (V k ) k∈N is compatible if and only if it satisfies the following two sets of properties: (2)

Suitable families of open subsets of the A k 's
Our starting point is the following very basic lemma on simplicial sets: (i) The simplex σ * (x) belongs to τ * (A m ).
(ii) There is an epimorphism ρ : Proof. The only non-trivial statement is that (i) implies (ii). Assume then that σ * (x) = τ * (y) for some y ∈ A m . We may choose a section δ : If τ • δ were not one-to-one, we would be able to decompose it as τ • δ = σ ′ • s n i for some i and some morphism σ ′ , which would yield x ∈ s i (A n−1 ), contradicting the fact that x is non-degenerate.
The following properties then generalize the axioms defining an x-admissible family: (ii) Let x ′ ∈ U σ and τ : Clearly, I σ is non-empty and better, for every (i, j) ∈ [k] 2 such that σ(i) = σ(j), there is some δ ∈ I σ with i and j in its range.
The U (σ) sets have the following main properties: Proof. (a) trivially derives from statement (i) in Lemma 2.2 and the definition of I σ .
. By a previous remark, there is some δ ∈ I σ the range of which contains y and z. Hence τ (y) = τ (z).

The category Γ ′
The relation ≤ on P(N) defined by yields a structure of poset 1 on P(N). We define the category Γ ′ as the one with the same objects as ∆ and for which, for any (k, k ′ ) ∈ N 2 , the morphisms from [k] to [k ′ ] are the increasing maps P([k]) → P([k ′ ]) which respect disjoint unions and map non-empty sets to non-empty sets, with the obvious composition of morphisms. To avoid any confusion with the simplicial category, a morphism f To any such morphism corresponds an epimorphism σ : .

The family W (f, ε) of open subsets of the simplicies
.
Clearly, such a family exists, and we may choose one for the rest of the section. Let then ε ∈ ]0, 1[. For any f : Obviously, this is an open convex subset of ∆ k and its closure is the set of those Notice finally that α ∈ W (id [m] , ε).

Completing the construction of
Given an onto morphism f : , thus corresponding to an epimorphism σ : Obviously, U (f ) is an open subset of A k .

Ordering the morphisms of
Assume now that i < k ′ . If # f (i) = 0 and # f (i + 1) ≥ 1, we define f +i by: If # f (i) ≥ 1 and # f (i + 1) = 0, we define f +i by: In any case, f +i is obtained from f by attaching i or i + 1 to an adjacent set of the form f {k}. We denote by R the binary relation defined on by f Rf +i for every i and every f for which f +i is defined. We then define ≤ as the pre-order relation generated by R. Actually, this is an order relation on Notice that, whenever f +i is defined, f ⊂ f +i . It follows that f ≤ g ⇒ f ⊂ g for every f and g in Hom is a poset (its maximal elements are the onto morphisms). The opposite order relation will be denoted by ≥. , one obviously has f ⊂ g whilst the statement f ≤ g is false (notice that f ⊂ g is an irreducible chain for ⊂ whereas no i satisfies g = f +i ).

The definition of U ε
Let ε ∈ ]0, 1[ and k ∈ N. Set then which is clearly an open subset of A k × ∆ k . Set also and notice that U k,ε ⊂ U ′ k,ε . For an arbitrary ε ∈ ]0, 1[, we finally define: For every ε ∈]0, 1[, one has α ∈ W (id [m] , ε) and x ∈ U (id [m] ), hence (x, α) ∈ U m,ε since id [m] is maximal. This shows that [(x, α)] ∈ U ε . In the next section, we will show that U ε is an open subset of |A| by proving that the family (U k,ε ) k∈N is compatible.

The proof that U ε is an open subset of |A|
Here, we will prove the following proposition: Proposition 3.1. Let ε ∈ ]0, 1[. Then: (a) For every k ∈ N * , (y, β) ∈ A k × ∆ k−1 and i ∈ [k]: This has the following immediate corollary as explained in the introduction of Section 2:

One last notation
.

Proof of statement (a) in Proposition 3.1
We fix an arbitrary pair (y, β) ∈ A k × ∆ k−1 and an arbitrary integer i ∈ [k]. Assume first that there exists some f : .
The rest of the proof essentially rests upon the following claim: Notice conversely that, for an arbitrary z, the condition the epimorphisms corresponding to red(f ) and red(f −i ). Then there is some j such that σ • δ j = σ ′ and the square is commutative. However d j (sup(f ) * (y)) ∈ U (σ ′ ) since sup(f ) * (y) ∈ U (σ) and δ j ∈ I σ (cf. statement (c) in Proposition 2.3). It follows that sup(f −i ) * (d i (y)) ∈ U (σ ′ ) hence d i (y) ∈ U (f −i ).
• Let now g ≥ f −i . We wish to prove that β ∈ W (g, ε).

Conversely, assume
It was proven earlier that if # g (i) = 0, then, for every z, one has z ∈ U (f ) if and only if d i (z) ∈ U (f −i ). Applying this to (δ i ) * (f ) yields y ∈ U ((δ i ) * (f )) since (δ i ) * (f ) −i = f . On the other hand, for any g ≥ (δ i ) * (f ), Lemma 3.3 shows and then (δ i ) * (β) ∈ W (g, ε) by using the definition of the W (h, η)'s. We deduce that which finishes the proof of point (a) in Proposition 3.1.
This finishes the proof of Proposition 3.1.
Remark 5. One might wonder whether we could have avoided all those technicalities, especially the introduction of ≤, and why not simply define U k,ε as U (f ) × W (f, ε) (which was our first idea). The trouble is that the last points in the proof of statement (b) of Proposition 3.1 miserably fail if such a definition is considered.

Application to simplicial Hausdorff spaces
In this section, we assume that A is a simplicial Hausdorff space. We arbitrarily choose two pairs (x, α) and (y, β) such that x ∈ A n is non-degenerate, y ∈ A m is non degenerate, α ∈ ∆ n ∂∆ n and β ∈ ∆ m ∂∆ m . We also pick an arbitrary integer N ≥ max(n, m). . Using the procedure of Sections 2.2 and 2.1, we recover two families (U (f )) and (V (g)).
For every k ∈ N, the procedure of Section 2.5.3 yields subsets U k,ε and U ′ k,ε of A k × ∆ k from (U (f )) and (W (f, ε)), and subsets V k,ε and V ′ k,ε of A k × ∆ k from (V (g)) and (T (g, ε)). This yields open subsets U ε and V ε of |A|.
4.2 Finishing the proof of Theorem 1.1 Theorem 1.1 is now within our reach. We now assume (x, α) = (y, β). We will then show that the various admissible families that were used in the construction of the open subset U ε and V ε may be carefully chosen so that the assumptions of Proposition 4.1 are satisfied. We will need to tackle separately the case x = y and the case x = y. The following basic lemma on simplicial sets will prove essential:  such that σ * (x) = τ * (y). Applying lemma 2.1 to both x and y, we easily find that m = n and σ = τ . However, σ * is one-to-one since σ has a section in ∆. Hence x = y, which proves both statements in the lemma.
The following lemma will then show that we may use Proposition 4.1, which will complete the case x = y:
Proposition 4.1 may then be used thanks to the next lemma, and this will complete both the case x = y and the proof of Theorem 1.1: then σ = τ . Thus σ • σ ′ = τ • σ ′ and we deduce from the hypothesis that U σ•σ ′ ∩ U τ •σ ′ = ∅, which yields U σ ∩ U τ = ∅.
Indeed, it is obvious on the one hand that (σ l ) * (γ) ∈ W (f −l , ε) ∩ T (g −l , ε); on the other hand σ • δ l and τ • δ l clearly are the epimorphisms respectively associated to the onto morphisms f −l and g −l , hence point (c) in Lemma 2.3 shows d l (z) ∈ U (f −l ) ∩ U (g −l ). Finally, the contradiction comes from the induction hypothesis since f −l and g −l are both onto.