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\begin{document}

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\section*{colimits of normal spaces}

\hypertarget{context}{}\subsubsection*{{Context}}\label{context}

\hypertarget{limits_and_colimits}{}\paragraph*{{Limits and colimits}}\label{limits_and_colimits}

[[!include infinity-limits - contents]]

\hypertarget{topology}{}\paragraph*{{Topology}}\label{topology}

[[!include topology - contents]]

\hypertarget{contents}{}\section*{{Contents}}\label{contents}

\noindent\hyperlink{basic_results}{Basic results}\dotfill \pageref*{basic_results} \linebreak
\noindent\hyperlink{tietze_characterization}{Tietze characterization}\dotfill \pageref*{tietze_characterization} \linebreak
\noindent\hyperlink{sequences_of_normal_spaces}{Sequences of normal spaces}\dotfill \pageref*{sequences_of_normal_spaces} \linebreak
\noindent\hyperlink{related_entries}{Related entries}\dotfill \pageref*{related_entries} \linebreak


This page will collect some technical material concerning [[colimits]] of [[normal spaces]] as computed in [[Top]]. Here ``normal'' means $T_4$, i.e., we assume as part of normality that points are closed.

In particular it shows that [[CW-complexes are paracompact Hausdorff spaces|CW-complexes are normal spaces]] (theorem \ref{CWComplexesAreNormal} 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]].

\hypertarget{basic_results}{}\subsection*{{Basic results}}\label{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

\begin{displaymath}
\mathbb{R} \leftarrow \mathbb{R} \setminus \{0\} \to \mathbb{R}
\end{displaymath}
where both maps are the inclusion map. This space isn't even [[Hausdorff space|Hausdorff]], as every [[neighborhood]] of the point $0_1$ (the image of the origin under the first pushout [[coprojection]]) intersects every neighborhood of the point $0_2$ (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 $T_1$, i.e., that [[singletons]] are [[closed set|closed]] (\href{separation+axioms#T1InTermsOfClosureOfPoints}{here}), so that ``normal'' means $T_4$).

\begin{prop}
\label{normal}\hypertarget{normal}{}
Let $X$ be normal, and suppose $q: X \to Y$ in $Top$ is a [[closed maps|closed]] [[surjective function|surjection]]. Then $Y$ is normal.

\end{prop}
\begin{proof}
First we claim singletons of $Y$ are closed: for each $y \in Y$ there exists $x \in X$ such that $y = q(x)$. Since $\{x\}$ is closed in $X$ and $q$ is closed, $\{y\} = q(\{x\})$ is closed in $Y$.

Let $C, D$ be disjoint closed sets of $Y$, so that $\neg C, \neg D$ form an open cover of $Y$. Then $A = q^{-1}(\neg C), B = q^{-1}(\neg D)$ form an open cover of $X$. Normality of $X$ (in ``[[De Morgan duality|De Morganized]]'' form) implies there are closed subsets $E \subseteq A, F \subseteq B$ which together also cover $X$. Then $Y = q(X) = q(E \cup F) = q(E) \cup q(F)$, so the closed sets $q(E), q(F)$ cover $Y$. And we have $q(E) \subseteq q(A) = \neg C$ and similarly $q(F) \subseteq \neg D$, which is equivalent to $C \subseteq \neg q(E)$ and $D \subseteq \neg q(F)$, where $\neg q(E), \neg q(F)$ are disjoint open sets of $Y$, and we are done.

\end{proof}
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.)

\begin{lemma}
\label{pretopos}\hypertarget{pretopos}{}
Let $X$ be a compact Hausdorff space. Then an equivalence relation $\sim \hookrightarrow X \times X$ is closed as a subset of $X \times X$ iff the quotient map $q: X \to X/\sim$ is a closed map. (In which case $X/\sim$ is compact Hausdorff.)

\end{lemma}
\begin{proof}
If $q: X \to X/\sim$ is a closed map, making $X/\sim$ compact Hausdorff, then the diagonal $\Delta \hookrightarrow X\sim \times X/\sim$ is a [[embedding of topological spaces|closed embedding]], so that its pullback along $q \times q$

\begin{displaymath}
\itexarray{
\sim & \hookrightarrow & X \times X \\ 
\downarrow & & \downarrow \mathrlap{q \times q} \\ 
\Delta & \hookrightarrow & X/{\sim} \times X/{\sim}
}
\end{displaymath}
also defines a closed subset of $X \times X$.

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

\begin{displaymath}
\{x \in X: (\exists_{a \in A})\; x\sim a\} = \pi_1((\sim)\cap (X \times A)).
\end{displaymath}
This is a closed set because the first projection $\pi_1$ is a closed map (by compactness of $X$). Moreover, $\bar{A} = q^{-1}(q(A))$, essentially by definition ($x \in q^{-1}(q(A))$ means $q(x) = q(a)$ for some $a \in A$). Since $q^{-1}(q(A))$ is closed, $q(A)$ is closed by definition of quotient topology.

\end{proof}
\begin{remark}
\label{}\hypertarget{}{}
A [[closed map|closed]] [[surjection]] is a quotient map (a [[regular epimorphism|regular epi]] in [[Top]]), and a closed [[injection]] is an [[embedding of topological spaces|embedding]] (a [[regular monomorphism|regular mono]] in $Top$).

\end{remark}
\begin{lemma}
\label{poclosed}\hypertarget{poclosed}{}
In $Top$, the pushout of a closed embedding along any continuous map is again a closed embedding.

\end{lemma}
\begin{proof}
We reproduce the proof given \href{/nlab/show/subspace+topology#pushout}{here}.

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

To prove that $j$ is a subspace, let $U \subseteq C$ be any open set. Then there exists open $V \subseteq B$ such that $i^\ast(V) = f^\ast(U)$ because $i$ is a subspace inclusion. If $\chi_U \colon C \to \mathbf{2}$ and $\chi_V \colon B \to \mathbf{2}$ 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 $\chi_W \colon D \to \mathbf{2}$ which extends the pair of maps $\chi_U, \chi_V$. It follows that $j^{-1}(W) = U$, so that $j$ is a subspace inclusion.

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

\end{proof}
\hypertarget{tietze_characterization}{}\subsection*{{Tietze characterization}}\label{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:

\begin{theorem}
\label{}\hypertarget{}{}
A space $X$ is normal if and only if, for each closed subset $C \subseteq X$ and map $f: C \to \mathbb{R}$, there is an extension map $g: X \to \mathbb{R}$, i.e., a map whose restriction $g|_C$ coincides with $f$.

\end{theorem}
\begin{remark}
\label{}\hypertarget{}{}
\begin{enumerate}%
\item There are variations on the theorem in which stronger separation properties (such as perfect normality, or $T_6$) may be reformulated in terms of extension conditions.


\item We remark that the ``if'' part of the proof is very easy. If $A, B$ are closed disjoint subsets of $X$, then the closed subspace $C = A \cup B$ is the coproduct on $A, B$ in $Top$, and we may define a map $f: A \cup B \to \mathbb{R}$ to be the constant $0$ on $A$ and the constant $1$ on $B$. Let $g: X \to \mathbb{R}$ be any extension of $f$; then $U = g^{-1}(\{x: x \lt 1/3\})$ and $V = g^{-1}(\{x: x \gt 2/3\})$ are separating open sets.



\end{enumerate}
\end{remark}
The following is a sample application.

\begin{theorem}
\label{attach}\hypertarget{attach}{}
If $X, Y, Z$ are normal spaces and $h: X \to Z$ is a closed embedding and $f: X \to Y$ is a continuous map, then in the pushout diagram in $Top$

\begin{displaymath}
\itexarray{
X & \stackrel{h}{\to} & Z \\ 
\mathllap{f} \downarrow & & \downarrow \mathrlap{g} \\ 
Y & \underset{k}{\to} & W,
}
\end{displaymath}
the space $W$ is normal (and $k: Y \to W$ is a closed embedding, by the preceding Lemma).

\end{theorem}
\begin{proof}
By the Tietze characterization, it suffices to show that any map $\phi: C \to \mathbb{R}$ on a closed subspace $C$ of $W$ can be extended to a map $\psi$ on all of $W$. Pulling back $C \hookrightarrow W$ along $k$, we have a closed subspace $k^{-1}(C) \hookrightarrow Y$ and a composite map $k^{-1}(C) \stackrel{k|}{\to} C \stackrel{\phi}{\to} \mathbb{R}$; call it $\alpha: k^{-1}(C) \to \mathbb{R}$. By normality of $Y$, the map $\alpha: k^{-1}(C) \to \mathbb{R}$ extends to a map $\beta: Y \to \mathbb{R}$. We obtain a map $\beta f: X \to \mathbb{R}$.

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

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

\end{proof}
\hypertarget{sequences_of_normal_spaces}{}\subsection*{{Sequences of normal spaces}}\label{sequences_of_normal_spaces}

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

\end{prop}
\begin{proof}
Let $A, B$ be disjoint closed subsets of $X$, and for all $n$ put $A_n = X_n \cap A$, $B_n = X_n \cap B$. Working recursively, suppose given disjoint open sets $U_n, V_n$ such that $A_n \subseteq U_n$ and $B_n \subseteq V_n$. Since normality guarantees that we can refine further if necessary, i.e., find an open $O_n$ such that $A_n \subseteq O_n$ and $\widebar{O_n} \subseteq U_n$, we may assume that the closures $\widebar{U_n}$, $\widebar{V_n}$ are disjoint in $X_n$, as are their images in $X_{n+1}$ under the closed embedding $i_n$. The closed sets $i_n(\widebar{U_n}) \cup A_{n+1}$ and $i_n(\widebar{V_n}) \cup B_{n+1}$ are disjoint in $X_{n+1}$:

\begin{itemize}%
\item $A_{n+1} \cap B_{n+1} = \emptyset$ since $A \cap B = \emptyset$,


\item $i_n(\widebar{U_n}) \cap i_n(\widebar{V_n}) = i_n((\widebar{U_n} \cap \widebar{V_n}) = i_n(\emptyset) = \emptyset$ where the direct image operator $i_n(-)$ preserves binary intersections since $i_n$ is monic,


\item $i_n(\widebar{U_n}) \cap B_{n+1} = i_n(\widebar{U_n} \cap i_n^{-1}(B_{n+1})) = i_n(\widebar{U_n} \cap B_n) \subseteq i_n((\widebar{U_n} \cap \widebar{V_n}) = \emptyset$ (using Frobenius reciprocity), and similarly,


\item $A_{n+1} \cap i_n(\widebar{V_n}) = \emptyset$.



\end{itemize}
Use normality of $X_{n+1}$ to select disjoint open sets $U_{n+1}, V_{n+1}$ such that $i_n(\widebar{U_n}) \cup A_{n+1} \subseteq U_{n+1}$ and $i_n(\widebar{V_n}) \cup B_{n+1} \subseteq V_{n+1}$, thus completing the recursive construction.

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

\end{proof}
\begin{remark}
\label{}\hypertarget{}{}
An alternative proof using the Tietze characterization is easily given: if $C \subseteq colim_n X_n$ is closed and $f: C \to \mathbb{R}$ is continuous, then putting $C_n = X_n \cap C$ and $f_n = f|_{C_n}$, we define a suitable extension of each $f_n: C_n \to \mathbb{R}$ to a map $g_n: X_n \to \mathbb{R}$ by induction. Supposing at stage $n$ the map $g_n$ is given, we have a well-defined map $h_{n+1}: C_{n+1} \cup i_n(X_n) \to \mathbb{R}$ defined as $f_{n+1}$ on $C_{n+1}$ and as the composite $i_n(X_n) \cong X_n \stackrel{g_n}{\to} \mathbb{R}$ on $i_n(X_n)$. Then using normality of $X_{n+1}$, extend $h_{n+1}$ to a map $g_{n+1}: X_{n+1} \to \mathbb{R}$. Clearly $g_{n+1}$ extends $f_{n+1}$ and we have the compatibility equations $g_n = g_{n+1} i_n$, so the $g_n$ paste together to form a map $g: colim_n X_n \to \mathbb{R}$ which extends $f$.

A little use of notation allows a short exposition of this proof. Let $Z$ be the topological space corresponding to the partial order $a\lt b\gt c\lt d\gt e$ (where $a,c,d$ are closed points, $b,d$ are open, and closed subsets are $\{a,b,c\},\{c,d,e\},\{a\},\{c\},\{e\},\emptyset$), and let $g$ be the map gluing together points $b,c,d$. By Theorem 2.1, $f$ has the left lifting property wrt $g$ whenever $f$ is a closed embedding into a normal space. As each $i_n: X_n \to X_{n+1}$, ${n \in \mathbb{N}}$ has the left lifting property wrt $g$, their transfinite composition $\emptyset\to colim_n X_n$ has the same lifting property, which means that $X$ is normal.

\end{remark}
\begin{theorem}
\label{CWComplexesAreNormal}\hypertarget{CWComplexesAreNormal}{}
A [[CW-complex]] is a [[normal space]].

\end{theorem}
\begin{proof}
A CW-complex $X$ is formed by an inductive process where the $n$-skeleton $X_n$ is formed as an [[attachment space]] formed from normal spaces. That is, we start with the normal space $X_{-1} = \emptyset$, and given the normal space $X_{n-1}$ and an attaching map $f: S_{n-1} = \sum_{i \in I} S_i^{n-1} \to X_{n-1}$, we push out the closed embedding $S_{n-1} = \sum_{i \in I} S_i^{n-1} \hookrightarrow \sum_{i \in I} D_i^n = D_n$ along the attaching map $f$ to get a closed embedding

\begin{displaymath}
i_n: X_{n-1} \to X_n (= X_{n-1} \cup_{S_{n-1}} D_n)
\end{displaymath}
and deduce $X_n$ is normal by Theorem \ref{attach}. Then $X = colim_n X_n$ is normal by applying Proposition \ref{sequence}.

\end{proof}
\hypertarget{related_entries}{}\subsection*{{Related entries}}\label{related_entries}

\begin{itemize}%
\item [[colimits of paracompact Hausdorff spaces]]

\end{itemize}


\end{document}