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colimits of normal spaces

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Limits and colimits

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topology (point-set topology, point-free topology)

see also differential topology, algebraic topology, functional analysis and topological homotopy theory

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Contents

This page will collect some technical material concerning colimits of normal spaces as computed in Top.

In particular it shows that CW-complexes are normal spaces (theorem 3 below).

A basic technique is the exploitation of the Tietze extension condition which characterizes normal spaces. Similar techniques may be used to prove a parallel set of results for paracompact spaces; see colimits of paracompact spaces.

Basic results

We start with some easy but useful results, using only our “bare hands” (i.e., using the definition of normality and applying simple reasoning).

The first very easy observation is that normal spaces are closed under coproducts in Top (so-called “disjoint union spaces”). The proof may be safely left to the reader.

There is no hope that normal spaces are closed under coequalizers or pushouts. A first example that comes to mind is the line with double origin, which is the topological pushout of the diagram

{0}\mathbb{R} \leftarrow \mathbb{R} \setminus \{0\} \to \mathbb{R}

where both maps are the inclusion map. This space isn’t even Hausdorff, as every neighborhood of the point 0 10_1 (the image of the origin under the first pushout coprojection) intersects every neighborhood of the point 0 20_2 (the image of the origin under the second coprojection).

However, there are reasonable conditions under which pushouts will be normal. (Throughout we assume all spaces are T 1T_1, i.e., that singletons are closed (here), so that “normal” means T 4T_4).

Proposition

Let XX be normal, and suppose q:XYq: X \to Y in TopTop is a closed surjection. Then YY is normal.

Proof

First we claim singletons of YY are closed: for each yYy \in Y there exists xXx \in X such that y=q(x)y = q(x). Since {x}\{x\} is closed in XX and qq is closed, {y}=q({x})\{y\} = q(\{x\}) is closed in YY.

Let C,DC, D be disjoint closed sets of YY, so that ¬C,¬D\neg C, \neg D form an open cover of YY. Then A=q 1(¬C),B=q 1(¬D)A = q^{-1}(\neg C), B = q^{-1}(\neg D) form an open cover of XX. Normality of XX (in “De Morganized” form) implies there are closed subsets EA,FBE \subseteq A, F \subseteq B which together also cover XX. Then Y=q(X)=q(EF)=q(E)q(F)Y = q(X) = q(E \cup F) = q(E) \cup q(F), so the closed sets q(E),q(F)q(E), q(F) cover YY. And we have q(E)q(A)=¬Cq(E) \subseteq q(A) = \neg C and similarly q(F)¬Dq(F) \subseteq \neg D, which is equivalent to C¬q(E)C \subseteq \neg q(E) and D¬q(F)D \subseteq \neg q(F), where ¬q(E),¬q(F)\neg q(E), \neg q(F) are disjoint open sets of YY, and we are done.

Here is one way in which such closed surjections arise: we know that compact Hausdorff spaces are normal spaces, and this is an especially nice class because the category of compact Hausdorff spaces is a pretopos. This is the categorical backdrop for the following observation. (We don’t need this lemma in any result below; we include it merely as a handy lemma to have around.)

Lemma

Let XX be a compact Hausdorff space. Then an equivalence relation X×X\sim \hookrightarrow X \times X is closed as a subset of X×XX \times X iff the quotient map q:XX/q: X \to X/\sim is a closed map. (In which case X/X/\sim is compact Hausdorff.)

Proof

If q:XX/q: X \to X/\sim is a closed map, making X/X/\sim compact Hausdorff, then the diagonal ΔX×X/\Delta \hookrightarrow X\sim \times X/\sim is a closed embedding, so that its pullback along q×qq \times q

X×X q×q Δ X/×X/\array{ \sim & \hookrightarrow & X \times X \\ \downarrow & & \downarrow \mathrlap{q \times q} \\ \Delta & \hookrightarrow & X/{\sim} \times X/{\sim} }

also defines a closed subset of X×XX \times X.

In the other direction, suppose \sim is closed, and let AXA \subseteq X be a closed subset of XX. Consider its \sim-saturation A¯\bar{A}, namely

{xX:( aA)xa}=π 1(()(X×A)).\{x \in X: (\exists_{a \in A})\; x\sim a\} = \pi_1((\sim)\cap (X \times A)).

This is a closed set because the first projection π 1\pi_1 is a closed map (by compactness of XX). Moreover, A¯=q 1(q(A))\bar{A} = q^{-1}(q(A)), essentially by definition (xq 1(q(A))x \in q^{-1}(q(A)) means q(x)=q(a)q(x) = q(a) for some aAa \in A). Since q 1(q(A))q^{-1}(q(A)) is closed, q(A)q(A) is closed by definition of quotient topology.

Remark

A closed surjection is a quotient map (a regular epi in Top), and a closed injection is an embedding (a regular mono in TopTop).

Lemma

In TopTop, the pushout of a closed embedding along any continuous map is again a closed embedding.

Proof

We reproduce the proof given here.

Since U=hom(1,):TopSetU = \hom(1, -): Top \to Set is faithful, we have that monos are reflected by UU; also monos and pushouts are preserved by UU since UU has both a left adjoint and a right adjoint. In SetSet, the pushout of a mono along any map is a mono, so we conclude jj is monic in TopTop. Furthermore, such a pushout diagram in SetSet is also a pullback, so that we have the Beck-Chevalley equality if *=g * j:P(C)P(B)\exists_i \circ f^\ast = g^\ast \exists_j \colon P(C) \to P(B) (where i:P(A)P(B)\exists_i \colon P(A) \to P(B) is the direct image map between power sets, and f *:P(C)P(A)f^\ast: P(C) \to P(A) is the inverse image map).

To prove that jj is a subspace, let UCU \subseteq C be any open set. Then there exists open VBV \subseteq B such that i *(V)=f *(U)i^\ast(V) = f^\ast(U) because ii is a subspace inclusion. If χ U:C2\chi_U \colon C \to \mathbf{2} and χ V:B2\chi_V \colon B \to \mathbf{2} are the maps to Sierpinski space that classify these open sets, then by the universal property of the pushout, there exists a unique continuous map χ W:D2\chi_W \colon D \to \mathbf{2} which extends the pair of maps χ U,χ V\chi_U, \chi_V. It follows that j 1(W)=Uj^{-1}(W) = U, so that jj is a subspace inclusion.

If moreover ii is an open inclusion, then for any open UCU \subseteq C we have that j *( j(U))=Uj^\ast(\exists_j(U)) = U (since jj is monic) and (by Beck-Chevalley) g *( j(U))= i(f *(U))g^\ast(\exists_j(U)) = \exists_i(f^\ast(U)) is open in BB. By the definition of the topology on DD, it follows that j(U)\exists_j(U) is open, so that jj is an open inclusion. The same proof, replacing the word “open” with the word “closed” throughout, shows that the pushout of a closed inclusion ii is a closed inclusion jj.

Tietze characterization

A powerful tool for proving theorems about topological colimits of normal spaces is the characterization of normal spaces via the Tietze extension theorem:

Theorem

A space XX is normal if and only if, for each closed subset CXC \subseteq X and map f:Cf: C \to \mathbb{R}, there is an extension map g:Xg: X \to \mathbb{R}, i.e., a map whose restriction g| Cg|_C coincides with ff.

Remarks
  1. There are variations on the theorem in which stronger separation properties (such as perfect normality, or T 6T_6) may be reformulated in terms of extension conditions.

  2. We remark that the “if” part of the proof is very easy. If A,BA, B are closed disjoint subsets of XX, then the closed subspace C=ABC = A \cup B is the coproduct on A,BA, B in TopTop, and we may define a map f:ABf: A \cup B \to \mathbb{R} to be the constant 00 on AA and the constant 11 on BB. Let g:Xg: X \to \mathbb{R} be any extension of ff; then U=g 1({x:x<1/3})U = g^{-1}(\{x: x \lt 1/3\}) and V=g 1({x:x>2/3})V = g^{-1}(\{x: x \gt 2/3\}) are separating open sets.

The following is a sample application.

Theorem

If X,Y,ZX, Y, Z are normal spaces and h:XZh: X \to Z is a closed embedding and f:XYf: X \to Y is a continuous map, then in the pushout diagram in TopTop

X h Z f g Y k W,\array{ X & \stackrel{h}{\to} & Z \\ \mathllap{f} \downarrow & & \downarrow \mathrlap{g} \\ Y & \underset{k}{\to} & W, }

the space WW is normal (and k:YWk: Y \to W is a closed embedding, by the preceding Lemma).

Proof

By the Tietze characterization, it suffices to show that any map ϕ:C\phi: C \to \mathbb{R} on a closed subspace CC of WW can be extended to a map ψ\psi on all of WW. Pulling back CWC \hookrightarrow W along kk, we have a closed subspace k 1(C)Yk^{-1}(C) \hookrightarrow Y and a composite map k 1(C)k|Cϕk^{-1}(C) \stackrel{k|}{\to} C \stackrel{\phi}{\to} \mathbb{R}; call it α:k 1(C)\alpha: k^{-1}(C) \to \mathbb{R}. By normality of YY, the map α:k 1(C)\alpha: k^{-1}(C) \to \mathbb{R} extends to a map β:Y\beta: Y \to \mathbb{R}. We obtain a map βf:X\beta f: X \to \mathbb{R}.

Similarly, pulling CWC \hookrightarrow W back along gg, we have a composite map g 1(C)g|Cϕg^{-1}(C) \stackrel{g|}{\to} C \stackrel{\phi}{\to} \mathbb{R}. We may now define a map

γ:h(X)g 1(C)\gamma: h(X) \cup g^{-1}(C) \to \mathbb{R}

by γ(h(x))=βf(x)\gamma(h(x)) = \beta f(x) for h(x)h(X)h(x) \in h(X), and γ(z)=ϕ(g(z))\gamma(z) = \phi(g(z)) for zg 1(C)z \in g^{-1}(C). It is not hard to check that γ\gamma is well-defined (by commutativity of the pullback square) and is continuous (using the fact that h|:Xh(X)h|: X \to h(X) is a homeomorphism, since hh is an embedding), and that h(X)g 1(C)h(X) \cup g^{-1}(C) is a closed subset of ZZ (since hh is closed). By normality of ZZ, we may extend γ\gamma to a map δ:Z\delta: Z \to \mathbb{R}, and we have just observed that δh=βf\delta h = \beta f, so the pair (β,δ):Y+Z(\beta, \delta): Y + Z \to \mathbb{R} induces a unique map ψ:W\psi: W \to \mathbb{R} such that ψk=β\psi k = \beta and ψg=δ\psi g = \delta. Finally, the restriction of ψ\psi to CC is ϕ\phi, as required.

Sequences of normal spaces

Proposition

If (i n:X nX n+1) n(i_n: X_n \to X_{n+1})_{n \in \mathbb{N}} is a countable sequence of closed embeddings between normal spaces, then the colimit X=colim nX nX = colim_n X_n is also normal.

Proof

Let A,BA, B be disjoint closed subsets of XX, and for all nn put A n=X nAA_n = X_n \cap A, B n=X nBB_n = X_n \cap B. Working recursively, suppose given disjoint open sets U n,V nU_n, V_n such that A nU nA_n \subseteq U_n and B nV nB_n \subseteq V_n. Since normality guarantees that we can refine further if necessary, i.e., find an open O nO_n such that A nO nA_n \subseteq O_n and O n¯U n\widebar{O_n} \subseteq U_n, we may assume that the closures U n¯\widebar{U_n}, V n¯\widebar{V_n} are disjoint in X nX_n, as are their images in X n+1X_{n+1} under the closed embedding i ni_n. The closed sets i n(U n¯)A n+1i_n(\widebar{U_n}) \cup A_{n+1} and i n(V n¯)B n+1i_n(\widebar{V_n}) \cup B_{n+1} are disjoint in X n+1X_{n+1}:

  • A n+1B n+1=A_{n+1} \cap B_{n+1} = \emptyset since AB=A \cap B = \emptyset,

  • i n(U n¯)i n(V n¯)=i n((U n¯V n¯)=i n()=i_n(\widebar{U_n}) \cap i_n(\widebar{V_n}) = i_n((\widebar{U_n} \cap \widebar{V_n}) = i_n(\emptyset) = \emptyset where the direct image operator i n()i_n(-) preserves binary intersections since i ni_n is monic,

  • i n(U n¯)B n+1=i n(U n¯i n 1(B n+1))=i n(U n¯B n)i n((U n¯V n¯)=i_n(\widebar{U_n}) \cap B_{n+1} = i_n(\widebar{U_n} \cap i_n^{-1}(B_{n+1})) = i_n(\widebar{U_n} \cap B_n) \subseteq i_n((\widebar{U_n} \cap \widebar{V_n}) = \emptyset (using Frobenius reciprocity), and similarly,

  • A n+1i n(V n¯)=A_{n+1} \cap i_n(\widebar{V_n}) = \emptyset.

Use normality of X n+1X_{n+1} to select disjoint open sets U n+1,V n+1U_{n+1}, V_{n+1} such that i n(U n¯)A n+1U n+1i_n(\widebar{U_n}) \cup A_{n+1} \subseteq U_{n+1} and i n(V n¯)B n+1V n+1i_n(\widebar{V_n}) \cup B_{n+1} \subseteq V_{n+1}, thus completing the recursive construction.

It is clear that i n(U n)U n+1i_n(U_n) \subseteq U_{n+1} and the union U=colim nU nU = colim_n U_n defines an open set of XX (by definition of colimit topology), as does V=colim nV nV = colim_n V_n. The sets U,VU, V include A,BA, B respectively and are disjoint since any element they have in common must belong to U nU_n and V nV_n for sufficiently large nn, which is impossible. This completes the proof.

Remark

An alternative proof using the Tietze characterization is easily given: if Ccolim nX nC \subseteq colim_n X_n is closed and f:Cf: C \to \mathbb{R} is continuous, then putting C n=X nCC_n = X_n \cap C and f n=f| C nf_n = f|_{C_n}, we define a suitable extension of each f n:C nf_n: C_n \to \mathbb{R} to a map g n:X ng_n: X_n \to \mathbb{R} by induction. Supposing at stage nn the map g ng_n is given, we have a well-defined map h n+1:C n+1i n(X n)h_{n+1}: C_{n+1} \cup i_n(X_n) \to \mathbb{R} defined as f n+1f_{n+1} on C n+1C_{n+1} and as the composite i n(X n)X ng ni_n(X_n) \cong X_n \stackrel{g_n}{\to} \mathbb{R} on i n(X n)i_n(X_n). Then using normality of X n+1X_{n+1}, extend h n+1h_{n+1} to a map g n+1:X n+1g_{n+1}: X_{n+1} \to \mathbb{R}. Clearly g n+1g_{n+1} extends f n+1f_{n+1} and we have the compatibility equations g n=g n+1i ng_n = g_{n+1} i_n, so the g ng_n paste together to form a map g:colim nX ng: colim_n X_n \to \mathbb{R} which extends ff.

Theorem

A CW-complex is a normal space.

Proof

A CW-complex XX is formed by an inductive process where the nn-skeleton X nX_n is formed as an attachment space formed from normal spaces. That is, we start with the normal space X 1=X_{-1} = \emptyset, and given the normal space X n1X_{n-1} and an attaching map f:S n1= iIS i n1X n1f: S_{n-1} = \sum_{i \in I} S_i^{n-1} \to X_{n-1}, we push out the closed embedding S n1= iIS i n1 iID i n=D nS_{n-1} = \sum_{i \in I} S_i^{n-1} \hookrightarrow \sum_{i \in I} D_i^n = D_n along the attaching map ff to get a closed embedding

i n:X n1X n(=X n1 S n1D n)i_n: X_{n-1} \to X_n (= X_{n-1} \cup_{S_{n-1}} D_n)

and deduce X nX_n is normal by Theorem 2. Then X=colim nX nX = colim_n X_n is normal by applying Proposition 2.

Revised on June 6, 2017 04:57:32 by Urs Schreiber (88.77.226.246)