description/proof of that for 2 pointed continuous maps, wedge sum of maps is continuous
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of continuous map.
- The reader knows a definition of wedge sum of pointed topological spaces.
- The reader knows a definition of wedge sum of pointed maps.
- The reader admits the proposition that for any maps composition, the preimage under the composition is the composition of the map preimages in the reverse order.
Target Context
- The reader will have a description and a proof of the proposition that for any 2 pointed continuous maps, the wedge sum of the maps is continuous.
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:
\((T_1, p_1)\): \(\in \{\text{ the pointed topological spaces }\}\)
\((T_2, p_2)\): \(\in \{\text{ the pointed topological spaces }\}\)
\(T_1 \vee T_2\): \(= \text{ the wedge sum of the topological spaces }\)
\((T, p)\): \(\in \{\text{ the pointed topological spaces }\}\)
\(f_1\): \(T_1 \to T\), \(\in \{\text{ the pointed continuous maps such that } f_1 (p_1) = p\}\)
\(f_2\): \(T_2 \to T\), \(\in \{\text{ the pointed continuous maps such that } f_2 (p_2) = p\}\)
\(f_1 \vee f_2\): \(: T_1 \vee T_2 \to T\), \(= \text{ the wedge sum of the pointed maps }\)
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Statements:
\(f_1 \vee f_2 \in \{\text{ the continuous maps }\}\)
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2: Proof
Whole Strategy: Step 1: let \(\pi: T_1 + T_2 \to T_1 \vee T_2\) be the quotient map; Step 2: take any open subset, \(U \subseteq T\), set the goal of seeing that \((f_1 \vee f_2)^{-1} (U) \subseteq T_1 \vee T_2\) is open; Step 3: see that \(\pi^{-1} ((f_1 \vee f_2)^{-1} (U)) = {f_1}^{-1} (U) \cup {f_2}^{-1} (U)\); Step 4: see that \((f_1 \vee f_2)^{-1} (U) \subseteq T_1 \vee T_2\) is open.
Step 1:
\(\pi: T_1 + T_2 \to T_1 \vee T_2\) is the quotient map, where \(T_1 + T_2\) is the topological sum.
Step 2:
Let \(U \subseteq T\) be any open subset.
Our goal is to see that \((f_1 \vee f_2)^{-1} (U) \subseteq T_1 \vee T_2\) is open.
That is about whether \(\pi^{-1} ((f_1 \vee f_2)^{-1} (U)) \cap T_1 \subseteq T_1\) and \(\pi^{-1} ((f_1 \vee f_2)^{-1} (U)) \cap T_2 \subseteq T_2\) are open.
Step 3:
Let us prove that \(\pi^{-1} ((f_1 \vee f_2)^{-1} (U)) = {f_1}^{-1} (U) \cup {f_2}^{-1} (U)\).
Let \(p' \in {f_1}^{-1} (U) \cup {f_2}^{-1} (U)\) be any. \(p' \in {f_j}^{-1} (U)\) for a \(j\). \(f_j (p') \in U\).
When \(p' = p_j\), \(\pi (p') = [p_j]\). \((f_1 \vee f_2) (\pi (p')) = p = f_j (p') \in U\); so, \(p' \in ((f_1 \vee f_2) \circ \pi)^{-1} (U) = \pi^{-1} ((f_1 \vee f_2)^{-1} (U))\): the proposition that for any maps composition, the preimage under the composition is the composition of the map preimages in the reverse order.
When \(p' \neq p_j\), \(\pi (p') = p'\), \((f_1 \vee f_2) (\pi (p')) = f_j (p') \in U\); so, \(p' \in ((f_1 \vee f_2) \circ \pi)^{-1} (U) = \pi^{-1} ((f_1 \vee f_2)^{-1} (U))\): the proposition that for any maps composition, the preimage under the composition is the composition of the map preimages in the reverse order.
So, \(p' \in \pi^{-1} ((f_1 \vee f_2)^{-1} (U))\) anyway.
Let \(p' \in \pi^{-1} ((f_1 \vee f_2)^{-1} (U))\) be any. \(p' \in T_j\) for a \(j\). \((f_1 \vee f_2) (\pi (p')) \in U\).
When \(p' = p_j\), \(\pi (p') = [p_j]\), \((f_1 \vee f_2) (\pi (p')) = (f_1 \vee f_2) ([p_j]) = f_j (p_j) \in U\); so, \(p' = p_j \in {f_j}^{-1} (U)\).
When \(p' \neq p_j\), \(\pi (p') = p'\), \((f_1 \vee f_2) (\pi (p')) = (f_1 \vee f_2) (p') = f_j (p') \in U\); so, \(p' \in {f_j}^{-1} (U)\).
So, \(p' \in {f_1}^{-1} (U) \cup {f_2}^{-1} (U)\) anyway.
Step 4:
So, \(\pi^{-1} ((f_1 \vee f_2)^{-1} (U)) \cap T_j = ({f_1}^{-1} (U) \cup {f_2}^{-1} (U)) \cap T_j = {f_j}^{-1} (U) \cap T_j = {f_j}^{-1} (U)\), which is open on \(T_j\), because \(f_j\) is continuous.
By the definition of topological sum, \(\pi^{-1} ((f_1 \vee f_2)^{-1} (U)) \subseteq T_1 + T_2\) is open.
