420: 2 Continuous Maps from Connected Topological Space Such That, for Any Point, if They Agree at Point, They Agree on Neighborhood and if They Disagree at Point, They Disagree on Neighborhood, Totally Agree or Totally Disagree

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A description/proof of that 2 continuous maps from connected topological space such that, for any point, if they agree at point, they agree on neighborhood and if they disagree at point, they disagree on neighborhood, totally agree or totally disagree

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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: Description
• 2: Proof
• 3: Note

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Starting Context

• The reader knows a definition of connected topological space.
• The reader knows a definition of continuous map.
• The reader knows a definition of neighborhood of point.
• The reader admits the proposition that any topological space is connected if and only if its open and closed subsets are only the topological space and the empty set.

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Target Context

• The reader will have a description and a proof of the proposition that any 2 continuous maps from any connected topological space into any topological space such that, for any point, if they (the maps) agree at the point, they agree on a neighborhood and if disagree at the point they disagree on a neighborhood, totally agree or totally disagree on the whole domain.

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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: Description

For any connected topological space, $T_1$, any topological space, $T_2$, and any continuous maps, $f_1,f_2:T_1\to T_2$, such that, for any point, $p\in T_1$, if $f_1(p)=f_2(p)$, there is a neighborhood, $N_p$, of $p$ such that $f_1(p^{\prime })=f_2(p^{\prime })$ for each $p^{\prime }\in N_p$ and if $f_1(p)\ne f_2(p)$, there is a neighborhood, $N_p$, of $p$ such that $f_1(p^{\prime })\ne f_2(p^{\prime })$ for each $p^{\prime }\in N_p$, $f_1(p)=f_2(p)$ for each $p\in T_1$ or $f_1(p)\ne f_2(p)$ for each $p\in T_1$.

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2: Proof

Such neighborhoods can be chosen to be open neighborhoods, because any neighborhood on which $f_1$ and $f_2$ agree or disagree contains an open neighborhood, on which $f_1$ and $f_2$ agree or disagree, respectively. So, there is an open neighborhood, $U_p$, around each point, $p\in T_1$, such that $f_1$ and $f_2$ agree or disagree on $U_p$.

Let us define $U:={\cup }_{p\in A}{U}_p$ where $A=\{p\in T_1|f_1(p)=f_2(p)\}$, the set of the points at which $f_1$ and $f_2$ agree. In fact, $U=A$, because for any point, $p\in A$, $p\in U$, and for any point, $p\in U$, $p\in A$. $U$ is open. Likewise, let us define $U^{\prime }:={\cup }_{p\in A^{\prime }}{U}_p$ where $A^{\prime }=\{p\in T_1|f_1(p)\ne f_2(p)\}$, the set of the points at which $f_1$ and $f_2$ disagree. In fact, $U^{\prime }=A^{\prime }$, because for any point, $p\in A^{\prime }$, $p\in U^{\prime }$, and for any point, $p\in U^{\prime }$, $p\in A^{\prime }$. $U^{\prime }$ is open.

In fact, $A=T_1\setminus A^{\prime }$, because each point, $p\in T_1$, belongs to exactly one of them. So, $A$ is closed. So, $A=T_1$ or $A=\mathrm{\varnothing }$, by the proposition that any topological space is connected if and only if its open and closed subsets are only the topological space and the empty set. If $A=T_1$, $f_1$ and $f_2$ agree on the whole $T_1$. If $A=\mathrm{\varnothing }$, $A^{\prime }=T_1$, and so, $f_1$ and $f_2$ disagree on the whole $T_1$.

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3: Note

It is required that $T_1$ is connected: otherwise, $f_1$ and $f_2$ can agree on a connected component of $T_1$ and disagree on another connected component of $T_1$.

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References

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