116: Some Properties About Adjunction Topological Space When Inclusion to Attaching-Origin Space from Subset Is Closed Embedding

<The previous article in this series | The table of contents of this series | The next article in this series>

A description/proof of some properties about adjunction topological space when inclusion to attaching-origin space from subset is closed embedding

---

Topics

About: topological space

---

The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

---

Starting Context

• The reader knows a definition of normal topological space.
• The reader knows a definition of adjunction topological space obtained by attaching topological space via continuous map to topological space.
• The reader knows a definition of continuous embedding.
• The reader knows a definition of closed map.
• The reader admits the proposition that for any adjunction topological space, the canonical map from the attaching-destination space to the adjunction space is a continuous embedding.
• The reader admits the proposition that for any map, the map preimage of any union of sets is the union of the map preimages of the sets.
• The reader admits the proposition that any restriction of any continuous map on the domain and the codomain is continuous.
• The reader admits the proposition that any closed set on any closed subspace topological space is closed on the base topological space.
• The reader admits the proposition that the preimages of any disjoint subsets under any map are disjoint.
• The reader admits the proposition that for any map between sets and its any domain-restriction, the preimage under the domain-restricted map is the intersection of the preimage under the original map and the restricted domain.
• The reader admits the proposition that any map between topological spaces is continuous if the domain restriction of the map to each closed set of a finite closed cover is continuous.
• The reader admits the Tietze extension theorem.
• The reader admits the reverse of Tietze extension theorem.
• The reader admits the proposition that for any adjunction topological space and its any subspace, any subset of the subspace is open if and only if the projections of the preimage of the subset under the quotient map onto the attaching-origin space and the attaching-destination space are open on the projections of the preimage of the subspace under the quotient map with the condition that the attaching-origin space projection accord with the attaching-destination space projection with respect to the attaching map.

---

Target Context

• The reader will have a description and a proof of some properties concerning any adjunction topological space when the inclusion to the attaching-origin space from the subset is a closed embedding.

---

Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

---

Main Body

---

1: Description

For any topological spaces, $T_1,T_2$, any subset, $S\subseteq T_1$, any continuous map, $f:S\to T_2$, the adjunction topological space, $T_2{\cup }_f{T}_1$, $f_1:S\to T_1$, $f_2:T_1\to T_2{\cup }_f{T}_1$, $f_3:T_2\to T_2{\cup }_f{T}_1$, and $f_4:T_1+T_2\to T_2{\cup }_f{T}_1$, when $f_1$ is a closed embedding, these properties hold: 1) $f_3$ is a closed embedding; 2) $f_2{|}_{T_1\setminus S}:T_1\setminus S\to T_2{\cup }_f{T}_1$ is an open embedding; 3) if $T_1$ and $T_2$ are spaces such that 3-1) each 1 point subset is closed or 3-2) $T_1$ and $T_2$ are normal, $T_2{\cup }_f{T}_1$ satisfies 3-1) or 3-2); 4) if $f$ is a quotient map, $f_2$ is a quotient map.

---

2: Proof

As for 1), by the proposition that for any adjunction topological space, the canonical map from the attaching-destination space to the adjunction space is a continuous embedding, $f_3$ is an embedding. For any closed set, $C\subseteq T_2$, $C=C^{\prime }\cup C^{″}$ where $C^{\prime }=C\cap f(S)$ and $C^{″}=C\setminus C^{\prime }$. $f_3(C)=f_3(C^{\prime })\cup f_3(C^{″})=\{[p]\in T_2{\cup }_f{T}_1|p\in C^{\prime },[p]=f^{-1}(p)\cup \{p\}\}\cup C^{″}$ where the last $C^{″}$ is of course not really exactly $C^{″}$ but the set of equivalent classes such that each equivalent class consists of the corresponding single point in $C^{″}$, which is $[p]=\{p\}$, but is denoted so because introducing another expression seems to complicate notations. Note that similar expressions are used henceforth.

Is $f_3(C)$ closed? By the definition of quotient topology, its closed-ness is nothing but the closed-ness of ${f_4}^{-1}(f_3(C))$, whose closed-ness is nothing but the closed-ness of ${f_4}^{-1}(f_3(C))\cap T_1$ and the closed-ness of ${f_4}^{-1}(f_3(C))\cap T_2$, by the definition of topological sum. But ${f_4}^{-1}(f_3(C))={f_4}^{-1}(\{[p]\in T_2{\cup }_f{T}_1|p\in C^{\prime },[p]=f^{-1}(p)\cup \{p\}\})\cup {f_4}^{-1}(C^{″})=f^{-1}(C^{\prime })\cup C^{\prime }\cup C^{″}=f^{-1}(C)\cup C$ by the proposition that for any map, the map preimage of any union of sets is the union of the map preimages of the sets. ${f_4}^{-1}(f_3(C))\cap T_1=f^{-1}(C)$ is closed on $S$ because $f$ is continuous; $f^{-1}(C)=f_1(f^{-1}(C))$ is closed on $T_1$ because $f_1$ is closed. ${f_4}^{-1}(f_3(C))\cap T_2=C$ is closed.

As for 2), $f_2$ is continuous, because $f_2:T_1\to T_1+T_2\to T_2{\cup }_f{T}_1$ is a compound of continuous maps (the former is an inclusion and the latter is $f_4$). By the proposition that any restriction of any continuous map on the domain and the codomain is continuous, $f_2{|}_{T_1\setminus S}$ is continuous, and is obviously injective. ${f_2{|}_{T_1\setminus S}}^{\prime }:T_1\setminus S\to f_2{|}_{T_1\setminus S}(T_1\setminus S)$ is obviously a continuous bijection. For any open set, $U\subseteq T_1\setminus S$, ${f_2{|}_{T_1\setminus S}}^{\prime }(U)=U$, which is open on $f_2{|}_{T_1\setminus S}(T_1\setminus S)$ if there is an open set, $U^{\prime }\subseteq T_2{\cup }_f{T}_1$, such that $U=U^{\prime }\cap f_2{|}_{T_1\setminus S}(T_1\setminus S)$, but $U^{\prime }=U$ will do, because $U$ is open on $T_2{\cup }_f{T}_1$, because the openness of $U$ is nothing but the openness of $f_4^{-1}(U)\cap T_1=U$ and the openness of $f_4^{-1}(U)\cap T_2=\mathrm{\varnothing }$, which hold true. So, ${f_2{|}_{T_1\setminus S}}^{\prime }$ is a homeomorphism, so, $f_2{|}_{T_1\setminus S}$ is an embedding. $f_2{|}_{T_1\setminus S}(U)=U$, which is open on $T_2{\cup }_f{T}_1$, as has been already shown. So, it is an open embedding.

As for 3) with 3-1), suppose that each 1 point subset on $T_1$ and on $T_2$ is closed. Any 1 point set on $T_2{\cup }_f{T}_1$ is $\{\{p\}\}$ where $p\in T_1\setminus S$ or $p\in T_2\setminus f(S)$ or $\{[p]\}$ where $p\in f(S)$ and $[p]=f^{-1}(p)\cup \{p\}$. The former is obviously closed. As for the latter, ${f_4}^{-1}([p])=f^{-1}(p)\cup \{p\}$, so, ${f_4}^{-1}([p])\cap T_1=f^{-1}(p)$ is closed on $S$ as $f$ is continuous, but as $f_1$ is closed, $S=f_1(S)$ is closed on $T_1$, so, ${f_4}^{-1}([p])\cap T_1$ is closed on $T_1$ by the proposition that any closed set on any closed subspace topological space is closed on the base topological space. ${f_4}^{-1}([p])\cap T_2=\{p\}$, closed.

As for 3) with 3-2), suppose that $T_1$ and $T_2$ are normal. By the reverse of Tietze extension theorem, if for any closed set, $C\subseteq T_2{\cup }_f{T}_1$, and any continuous map, $g:C\to \mathbb{R}$, there is a continuous extension, $\bar{g}:T_2{\cup }_f{T}_1\to \mathbb{R},\bar{g}{|}_C=g$, $T_2{\cup }_f{T}_1$ will be normal.

$S$ is closed on $T_1$, because $S=f_1(S)$ and $f_1$ is closed while $S$ is closed on $S$. ${f_2}^{-1}(C)={f_4}^{-1}(C)\cap T_1$ and ${f_3}^{-1}(C)={f_4}^{-1}(C)\cap T_2$ are closed by the definition of quotient topological space and the definition of topological sum.

1st, suppose that ${f_3}^{-1}(C)=T_2$, which implies that $S\subseteq {f_2}^{-1}(C)$. $h:{f_2}^{-1}(C)\to C,h=f_2{|}_{{f_2}^{-1}(C)}$, is continuous by the proposition that any restriction of any continuous map on the domain and the codomain is continuous. $g\circ h:{f_2}^{-1}(C)\to \mathbb{R}$ is continuous as a composition of continuous maps. As $T_1$ is normal, by the Tietze extension theorem, there is a continuous map, $i:T_1\to \mathbb{R},i{|}_{{f_2}^{-1}(C)}=g\circ h$. Let us define $\bar{g}$ by $\bar{g}\circ f_2=i$ on $f_2(T_1)$ and $\bar{g}\circ f_3=g\circ f_3$ on $f_3(T_2)$, which is well-defined because $f_2(T_1)$ and $f_3(T_2)$ cover $T_2{\cup }_f{T}_1$; for any $p\in T_1\setminus {f_2}^{-1}(C)$, $\bar{g}\circ f_2(p)=\bar{g}([p])=i(p)$, but $p\in T_1\setminus S$ as $S\subseteq {f_2}^{-1}(C)$, so, $\bar{g}(\{p\})=i(p)$, and for any $p\in {f_2}^{-1}(C)$, $\bar{g}\circ f_2(p)=\bar{g}([p])=i{|}_{{f_2}^{-1}(C)}(p)=g(h(p))=g([p])$; for any $p\in T_2$, $\bar{g}\circ f_3(p)=\bar{g}([p])=g\circ f_3(p)=g([p])$; $\bar{g}$ does not contradict for $S$ and for $f(S)$, because $\bar{g}=g$ there anyway.

Is $\bar{g}$ continuous? That is about that for any open set, $U\subseteq \mathbb{R}$, whether ${\bar{g}}^{-1}(U)$ is open. The openness is nothing but the openness of ${f_4}^{-1}({\bar{g}}^{-1}(U))\cap T_1={f_2}^{-1}({\bar{g}}^{-1}(U))$ and the openness of ${f_4}^{-1}({\bar{g}}^{-1}(U))\cap T_2={f_3}^{-1}({\bar{g}}^{-1}(U))$. The former is open because $\bar{g}\circ f_2=i$ is continuous. As for the latter, ${f_3}^{-1}({\bar{g}}^{-1}(U))={f_3}^{-1}({\bar{g}}^{-1}(U)\cap C)$ because $f_3$ maps into $C$ anyway by the supposition, ${f_3}^{-1}(C)=T_2$, but as $\bar{g}{|}_C=g$, ${\bar{g}}^{-1}(U)\cap C={\bar{g}{|}_C}^{-1}(U)=g^{-1}(U)$ by the proposition that for any map between sets and its any domain-restriction, the preimage under the domain-restricted map is the intersection of the preimage under the original map and the restricted domain. So, ${f_3}^{-1}({\bar{g}}^{-1}(U))={f_3}^{-1}(g^{-1}(U))$, which is open as a preimage of an open set under a composition of continuous maps.

Now, suppose that ${f_3}^{-1}(C)\subset T_2$. $j:{f_3}^{-1}(C)\to C,j=f_3{|}_{{f_3}^{-1}(C)}$, is continuous by the proposition that any restriction of any continuous map on the domain and the codomain is continuous. $g\circ j:{f_3}^{-1}(C)\to \mathbb{R}$ is continuous as a composition of continuous maps. As $T_2$ is normal, by the Tietze extension theorem, there is a continuous map, $k:T_2\to \mathbb{R},k{|}_{{f_3}^{-1}(C)}=g\circ j$. $f_3(T_2)$ is closed because ${f_4}^{-1}(f_3(T_2))\cap T_1=S$ and ${f_4}^{-1}(f_3(T_2))\cap T_2=T_2$ are closed on $T_1$ and $T_2$. Let us define $g^{\prime }:f_3(T_2)\cup C\to \mathbb{R}$ by $g^{\prime }\circ f_3=k$ on $f_3(T_2)$ and $g^{\prime }=g$ on $C$. It is well-defined because $f_3(T_2)$ and $C$ cover $f_3(T_2)\cup C$; for any $p\in T_2$, $g^{\prime }\circ f_3(p)=g^{\prime }([p])=k(p)$ but $p\mapsto [p]$ is injective; $g^{\prime }$ does not contradict on $f_3(T_2)\cap C$ because $k{|}_{{f_3}^{-1}(C)}=g\circ j$.

Is $g^{\prime }$ continuous? $\{f_3(T_2),C\}$ is a finite closed cover of $f_3(T_2)\cup C$, and if $g^{\prime }{|}_{f_3(T_2)}$ and $g^{\prime }{|}_C$ are continuous, $g^{\prime }$ will be continuous by the proposition that any map between topological spaces is continuous if the domain restriction of the map to each closed set of a finite closed cover is continuous. The latter is continuous, as it equals $g$. For the former, it is the matter of that for any $U\subseteq \mathbb{R}$, whether ${g^{\prime }{|}_{f_3(T_2)}}^{-1}(U)$ is open on $f_3(T_2)$ (remember that it is not on $T_2{\cup }_f{T}_1$). ${f_4}^{-1}({g^{\prime }{|}_{f_3(T_2)}}^{-1}(U))\cap T_1=f^{-1}({f_3}^{-1}({g^{\prime }{|}_{f_3(T_2)}}^{-1}(U)))=f^{-1}(k^{-1}(U))$, open on $S$ because $g^{\prime }{|}_{f_3(T_2)}\circ f_3=k$ is continuous and $f$ is continuous, which means that there is an open set, $U^{\prime }\subseteq T_1$ such that $U^{\prime }\cap S={f_4}^{-1}({g^{\prime }{|}_{f_3(T_2)}}^{-1}(U))\cap T_1$. $S={f_4}^{-1}(f_3(T_2))\cap T_1$. ${f_4}^{-1}({g^{\prime }{|}_{f_3(T_2)}}^{-1}(U))\cap T_2={f_3}^{-1}({g^{\prime }{|}_{f_3(T_2)}}^{-1}(U))=k^{-1}(U):=U^{″}$, open on $T_2$. $U^{\prime }\cap S=f^{-1}(U^{″})=f^{-1}(U^{″}\cap f(S))$. By the proposition that for any adjunction topological space and its any subspace, any subset of the subspace is open if and only if the projections of the preimage of the subset under the quotient map onto the attaching-origin space and the attaching-destination space are open on the projections of the preimage of the subspace under the quotient map with the condition that the attaching-origin space projection accord with the attaching-destination space projection with respect to the attaching map, as $U^{\prime }$ and $U^{″}$ satisfy the necessary conditions, ${g^{\prime }{|}_{f_3(T_2)}}^{-1}(U)$ is open on $f_3(T_2)$. So, $g^{\prime }$ is continuous.

Now, ${f_3}^{-1}(f_3(T_2)\cup C)=T_2$, the 1st case with $g^{\prime }$ instead of $g$, so, there is a continuous extension, $\overline{g^{\prime }}:T_2{\cup }_f{T}_1\to \mathbb{R}$, which is a continuous extension of $g$.

So, by the reverse of Tietze extension theorem, $T_2{\cup }_f{T}_1$ is normal.

As for 4), suppose that $f$ is a quotient map. $f_2$ is a continuous surjection. For any subset, $S^{\prime }\subseteq T_2{\cup }_f{T}_1$, suppose that ${f_2}^{-1}(S^{\prime })$ is open. Is $S^{\prime }$ open? The openness is nothing but the openness of ${f_4}^{-1}(S^{\prime })\cap T_1$ and the openness of ${f_4}^{-1}(S^{\prime })\cap T_2$. The former is nothing but ${f_2}^{-1}(S^{\prime })$, so, open. As for the latter, ${f^{-1}(f_4}^{-1}(S^{\prime })\cap T_2)={f_2}^{-1}(S^{\prime })\cap S$, which is open in the $S$ subspace topology as ${f_2}^{-1}(S^{\prime })$ is open on $T_1$, so, as $f$ is a quotient map, ${f_4}^{-1}(S^{\prime })\cap T_2$ is open.

---

References

<The previous article in this series | The table of contents of this series | The next article in this series>