859: For 2 Pointed Continuous Maps, Wedge Sum of Maps Is Continuous

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description/proof of that for 2 pointed continuous maps, wedge sum of maps is continuous

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Topics

About: topological space

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The table of contents of this article

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

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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.

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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.

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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.

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Main Body

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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 $p^{\prime }\in {f_1}^{-1}(U)\cup {f_2}^{-1}(U)$ be any. $p^{\prime }\in {f_j}^{-1}(U)$ for a $j$. $f_j(p^{\prime })\in U$.

When $p^{\prime }=p_j$, $\pi (p^{\prime })=[p_j]$. $(f_1\vee f_2)(\pi (p^{\prime }))=p=f_j(p^{\prime })\in U$; so, $p^{\prime }\in ((f_1\vee f_2)\circ \pi {)}^{-1}(U)={\pi }^{-1}((f_1\vee f_2{)}^{-1}(U))$: the proposition that for any maps composition, the preimage under the composition is the composition of the map preimages in the reverse order.

When $p^{\prime }\ne p_j$, $\pi (p^{\prime })=p^{\prime }$, $(f_1\vee f_2)(\pi (p^{\prime }))=f_j(p^{\prime })\in U$; so, $p^{\prime }\in ((f_1\vee f_2)\circ \pi {)}^{-1}(U)={\pi }^{-1}((f_1\vee f_2{)}^{-1}(U))$: the proposition that for any maps composition, the preimage under the composition is the composition of the map preimages in the reverse order.

So, $p^{\prime }\in {\pi }^{-1}((f_1\vee f_2{)}^{-1}(U))$ anyway.

Let $p^{\prime }\in {\pi }^{-1}((f_1\vee f_2{)}^{-1}(U))$ be any. $p^{\prime }\in T_j$ for a $j$. $(f_1\vee f_2)(\pi (p^{\prime }))\in U$.

When $p^{\prime }=p_j$, $\pi (p^{\prime })=[p_j]$, $(f_1\vee f_2)(\pi (p^{\prime }))=(f_1\vee f_2)([p_j])=f_j(p_j)\in U$; so, $p^{\prime }=p_j\in {f_j}^{-1}(U)$.

When $p^{\prime }\ne p_j$, $\pi (p^{\prime })=p^{\prime }$, $(f_1\vee f_2)(\pi (p^{\prime }))=(f_1\vee f_2)(p^{\prime })=f_j(p^{\prime })\in U$; so, $p^{\prime }\in {f_j}^{-1}(U)$.

So, $p^{\prime }\in {f_1}^{-1}(U)\cup {f_2}^{-1}(U)$ 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.

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References

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