description/proof of that for continuous map between topological spaces, image of connected subspace is connected
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of continuous, topological spaces map.
- The reader knows a definition of connected topological space.
- 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 the preimage of the whole codomain of any map is the whole domain.
- 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.
- The reader admits the proposition that the preimages of any disjoint subsets under any map are disjoint.
Target Context
- The reader will have a description and a proof of the proposition that for any continuous map between any topological spaces, the image of any connected subspace is connected.
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:
\(T'_1\): \(\in \{\text{ the topological spaces }\}\)
\(T'_2\): \(\in \{\text{ the topological spaces }\}\)
\(f'\): \(: T'_1 \to T'_2\), \(\in \{\text{ the continuous maps }\}\)
\(T_1\): \(\in \{\text{ the connected subspaces of } T'_1\}\)
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Statements:
\(f' (T_1) \in \{\text{ the connected subspaces of } T'_2\}\)
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2: Proof
Whole Strategy: Step 1: see that the domain and codomain restriction, \(f: T_1 \to f' (T_1)\), is continuous; Step 2: suppose that \(f' (T_1)\) was not connected, and see that \(T_1\) would not be connected.
Step 1:
Let us think of the domain and codomain restriction of \(f'\), \(f: T_1 \to f' (T_1), t \mapsto f' (t)\).
\(f\) is continuous, by the proposition that any restriction of any continuous map on the domain and the codomain is continuous.
Step 2:
Let us suppose that \(f' (T_1)\) was not connected as the subspace of \(T'_2\).
\(f' (T_1) = U_1 \cup U_2\), where \(U_1, U_2 \subseteq f' (T_1)\) would be nonempty open subsets of \(f' (T_1)\) such that \(U_1 \cap U_2 = \emptyset\).
\(f^{-1} (f' (T_1)) = T_1\), by the proposition that the preimage of the whole codomain of any map is the whole domain.
\(f^{-1} (f' (T_1)) = f^{-1} (U_1 \cup U_2) = f^{-1} (U_1) \cup f^{-1} (U_2)\), by the proposition that for any map, the map preimage of any union of sets is the union of the map preimages of the sets.
So, \(T_1 = f^{-1} (U_1) \cup f^{-1} (U_2)\).
\(f^{-1} (U_1)\) and \(f^{-1} (U_2)\) would be open on \(T_1\), because \(f\) was continuous.
\(f^{-1} (U_1) \neq \emptyset\), because \(U_1 \neq \emptyset\) while \(U_1 \subseteq f' (T_1) = f (T_1)\), so there would be a \(f (t) \in U_1\), and \(t \in f^{-1} (U_1)\); \(f^{-1} (U_2) \neq \emptyset\), likewise.
\(f^{-1} (U_1) \cap f^{-1} (U_2) = \emptyset\), by the proposition that the preimages of any disjoint subsets under any map are disjoint
So, \(T_1\) would not be connected, a contradiction.
So, \(f' (T_1)\) is connected as the subspace of \(T'_2\).
