definition of adjunction topological space obtained by attaching topological space via map to topological space
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of topological space.
- The reader knows a definition of continuous map.
- The reader knows a definition of topological sum.
- The reader knows a definition of quotient set.
- The reader knows a definition of quotient topology on set with respect to map.
Target Context
- The reader will have a definition of adjunction topological space obtained by attaching topological space via map to topological space.
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: Structured Description
Here is the rules of Structured Description.
Entities:
\( T_1\): \(\in \{\text{ the topological spaces }\}\)
\( T_2\): \(\in \{\text{ the topological spaces }\}\)
\( S\): \(\subseteq T_1\)
\( f\): \(: S \to T_2\), \(\in \{\text{ the continuous maps }\}\)
\( T_1 + T_2\): \(= \text{ the topological sum }\)
\( \sim\): \(\in \{\text{ the equivalence relations on } T_1 + T_2\}\), such that \(\forall t, t' \in T_1 + T_2 (t \sim t' \iff (t = t' \lor t' = f (t) \lor t = f (t') \lor f (t) = f (t')))\)
\( (T_1 + T_2) / \sim\): \(= \text{ the quotient set }\)
\( g: T_1 + T_2 \to (T_1 + T_2) / \sim, t \mapsto t' \in (T_1 + T_2) / \sim \text{ such that } t \in t'\)
\(*T_2 \cup_f T_1\): \(= (T_1 + T_2) / \sim\) with the quotient topology with respect to \(g\)
Conditions:
//
\(T_2 \cup_f T_1\) is called "adjunction topological space obtained by attaching \(T_1\) via \(f\) to \(T_2\)".
2: Note
Let us see that \(\sim\) is indeed an equivalence relation.
1) \(\forall t \in T_1 + T_2 (t \sim t)\): reflexivity: \(t = t\).
2) \(\forall t, t' \in T_1 + T_2 (t \sim t' \implies t' \sim t)\): symmetry: if \(t = t'\), \(t' = t\), so, \(t' \sim t\); if \(t' = f (t)\), \(t' \sim t\); if \(t = f (t')\), \(t' \sim t\); if \(f (t) = f (t')\), \(f (t') = f (t)\), so, \(t' \sim t\); so, anyway, \(t' \sim t\).
3) \(\forall t, t', t'' \in T_1 + T_2 ((t \sim t' \land t' \sim t'')\implies t \sim t'')\): transitivity: if \(t = t'\) or \(t' = t''\), \(t \sim t''\) is obvious, so, let us think of the other cases hereafter; if \(t' = f (t)\), the only possibility is \(t' = f (t'')\) (because \(t'' = f (t')\) is impossible, because \(t' \in T_2\); f (t') = f (t'') is impossible, because \(t' \in T_2\)), and as \(f (t) = t' = f (t'')\), \(t \sim t''\); if \(t = f (t')\), the only possibilities are \(t'' = f (t')\) and \(f (t') = f (t'')\) (because \(t' = f (t'')\) is impossible, because \(t' \in S\)), and if \(t'' = f (t')\), \(t = f (t') = t''\), so, \(t \sim t''\), and if \(f (t') = f (t'')\), \(t = f (t') = f (t'')\), so, \(t \sim t''\); if \(f (t) = f (t')\), the only possibilities are \(t'' = f (t')\) and \(f (t') = f (t'')\) (because \(t' = f (t'')\) is impossible, because \(t' \in S\)), and if \(t'' = f (t')\), \(t'' = f (t') = f (t)\), so, \(t \sim t''\); if \(f (t') = f (t'')\), \(f (t) = f (t') = f (t'')\), so, \(t \sim t''\); so, anyway, \(t \sim t''\).
\(g\) is valid, because for each \(t \in T_1 + T_2\), \(t' \in (T_1 + T_2) / \sim\) such that \(t \in t'\) is uniquely determined, because \(\sim\) is an equivalence relation.
\(g\) is indeed a surjection, because for each \(t' \in (T_1 + T_2) / \sim\), there is an element of \(t'\).
So, \(T_2 \cup_f T_1\) is valid.
