description/proof of that \(2\) constant maps from topological space into path-connected topological space are homotopic
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of path-connected topological space.
- The reader knows a definition of homotopic maps.
- The reader admits the proposition that any constant map between any topological spaces is continuous.
Target Context
- The reader will have a description and a proof of the proposition that any \(2\) constant maps from any topological space into any path-connected topological space are homotopic.
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:
\(T_1\): \(\in \{\text{ the topological spaces }\}\)
\(T_2\): \(\in \{\text{ the path-connected topological spaces }\}\)
\(p\): \(\in T_2\)
\(p'\): \(\in T_2\)
\(f\): \(: T_1 \to T_2, t \mapsto p\)
\(f'\): \(: T_1 \to T_2, t \mapsto p'\)
//
Statements:
\(f \simeq f'\), where \(\simeq\) means being homotopic
//
2: Proof
Whole Strategy: Step 1: see that \(f\) and \(f'\) are continuous; Step 2: take a path from \(p\) to \(p'\), \(\gamma: I \to T_2\), and a map, \(g: T_1 \times I \to T_2, (t, j) \mapsto \gamma (j)\), and see that \(g\) is a homotopy from \(f\) to \(f'\).
Step 1:
\(f\) and \(f'\) are continuous, by the proposition that any constant map between any topological spaces is continuous.
Step 2:
There is a path (which is a continuous map), \(\gamma: I \to T_2\), such that \(\gamma (0) = p\) and \(\gamma (1) = p'\), because \(T_2\) is path-connected.
Let us define the map, \(g: T_1 \times I \to T_2, (t, j) \mapsto \gamma (j)\).
\(g\) is continuous, because for each \((t, j) \in T_1 \times I\), for each neighborhood of \(g ((t, j)) = \gamma (j)\), \(N_{\gamma (j)} \subseteq T_2\), there is a neighborhood of \(j\), \(N_j \subseteq I\), such that \(\gamma (N_j) \subseteq N_{\gamma (j)}\), because \(\gamma\) is continuous, and there is the neighborhood of \((t, j)\), \(T_1 \times N_j \subseteq T_1 \times I\), which satisfies \(g (T_1 \times N_j) \subseteq N_{\gamma (j)}\), because for each \((t, j') \in T_1 \times N_j\), \(g ((t, j')) = \gamma (j') \in N_{\gamma (j)}\), because \(\gamma (N_j) \subseteq N_{\gamma (j)}\).
For each \(t \in T_1\), \(g ((t, 0)) = \gamma (0) = p = f (t)\) and \(g ((t, 1)) = \gamma (1) = p' = f ' (t)\).
So, \(f\) and \(f'\) are homotopic.
