193: For Adjunction Topological Space, Canonical Map from Attaching-Destination Space to Adjunction Space Is Continuous Embedding

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

A description/proof of that for adjunction topological space, canonical map from attaching-destination space to adjunction space is continuous 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 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 subspace topology.
• The reader knows a definition of quotient topology on set with respect to map.
• The reader knows a definition of topological sum.
• The reader admits the proposition that for any map, the map image of any union of sets is the union of the map images 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 for any map, the map preimage of any union of sets is the union of the map preimages of the sets.

---

Target Context

• The reader will have a description and a proof of the proposition that for any adjunction topological space, the canonical map from the attaching-destination space to the adjunction space is a continuous 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, $f:S\to T_2$, and the adjunction topological space, $T_2{\cup }_f{T}_1$, the canonical map, $f_3:T_2\to T_2{\cup }_f{T}_1$, is a continuous embedding.

---

2: Proof

$f_3$ is obviously injective. $f_3$ is continuous, because $f_3:T_2\to T_1+T_2\to T_2{\cup }_f{T}_1$ is a composition of continuous maps (the former constituent of the composition is so because of the definition of topological sum; the latter is so because of the definition of quotient space). ${f_3}^{\prime }:T_2\to f_3(T_2)$ is bijective and continuous by the proposition that any restriction of any continuous map on the domain and the codomain is continuous. The remaining issue is whether ${{f_3}^{\prime }}^{-1}$ is continuous. For any open set, $U\subseteq T_2$, $U=S^{\prime }\cup S^{″}$ where $S^{\prime }=U\cap f(T_1)$ and $S^{″}=U\setminus S^{\prime }$. Is ${{{f_3}^{\prime }}^{-1}}^{-1}(U)$ open on $f_3(T_2)$? ${{{f_3}^{\prime }}^{-1}}^{-1}(U)={f_3}^{\prime }(U)={f_3}^{\prime }(S^{\prime })\cup {f_3}^{\prime }(S^{″})$, by the proposition that for any map, the map image of any union of sets is the union of the map images of the sets. ${f_3}^{\prime }(S^{\prime })=\{[p]:p\in S^{\prime },[p]=\{f^{-1}(p)\cup \{p\}\}\}$; ${f_3}^{\prime }(S^{″})=S^{″}$.

Whether ${{{f_3}^{\prime }}^{-1}}^{-1}(U)$ is open on $f_3(T_2)$ is whether there is an open set, $U^{\prime }\subset T_2{\cup }_f{T}_1$, such that ${{{f_3}^{\prime }}^{-1}}^{-1}(U)=U^{\prime }\cap f_3(T_2)$, by the definition of subspace topology, so, let us define such a $U^{\prime }$. $f^{-1}(U)$ is open on $S$, so, there is an open set, $U^{″}\subseteq T_1$, such that $f^{-1}(U)=U^{″}\cap S$. $U^{\prime }:=(U^{″}\setminus S)\cup {{{f_3}^{\prime }}^{-1}}^{-1}(U)$. Let us prove that $U^{\prime }$ is open on $T_2{\cup }_f{T}_1$, which is whether $f_4^{-1}(U^{\prime })$ is open on $T_1+T_2$ where $f_4:T_1+T_2\to T_2{\cup }_f{T}_1$ is the quotient map, by the definition of quotient topology on set with respect to map, which is whether $f_4^{-1}(U^{\prime })\cap T_1$ and $f_4^{-1}(U^{\prime })\cap T_2$ are open, by the definition of topological sum. $f_4^{-1}(U^{\prime })\cap T_1=U^{″}$; $f_4^{-1}(U^{\prime })\cap T_2=U$. So, $U^{\prime }$ is open on $T_2{\cup }_f{T}_1$. ${{{f_3}^{\prime }}^{-1}}^{-1}(U)=U^{\prime }\cap f_3(T_2)$, because $U^{\prime }$ is $(U^{″}\setminus S)\cup {{{f_3}^{\prime }}^{-1}}^{-1}(U)$ while $f_3(T_2)\cap (U^{″}\setminus S)=\mathrm{\varnothing }$.

---

References

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