description/proof of that composition of homotopy equivalences is homotopy equivalence
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of homotopy equivalence.
- The reader admits the proposition that for any maps between any arbitrary subspaces of any topological spaces continuous at any corresponding points, the composition is continuous at the point.
- The reader admits the proposition that for any homotopic maps from any 1st topological space into any 2nd topological space and any homotopic maps from the 2nd topological space into any 3rd topological space, the compositions of the homotopic maps are homotopic with a homotopy as this.
Target Context
- The reader will have a description and a proof of the proposition that the composition of any homotopy equivalences is a homotopy equivalence.
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:
\(T_1\): \(\in \{\text{ the topological spaces }\}\)
\(T_2\): \(\in \{\text{ the topological spaces }\}\)
\(T_3\): \(\in \{\text{ the topological spaces }\}\)
\(f_1\): \(: T_1 \to T_2\), \(\in \{\text{ the homotopy equivalences }\}\)
\(f_2\): \(: T_2 \to T_3\), \(\in \{\text{ the homotopy equivalences }\}\)
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Statements:
\(f_2 \circ f_1 \in \{\text{ the homotopy equivalences }\}\)
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2: Note
Some people may think that this proposition is already included in the fact that being homotopy equivalent is an equivalence relation, but that fact implies only that there is a homotopy equivalence from \(T_1\) into \(T_3\), not \(f_2 \circ f_1\) is a homotopy equivalence: certainly, being an equivalence relation is usually proved by proving that \(f_2 \circ f_1\) is a homotopy equivalence, though.
3: Proof
Whole Strategy: Step 1: take a continuous \(\widetilde{f_2}: T_2 \to T_1\) such that \(\widetilde{f_2} \circ f_1 \simeq id_{T_1}\) and \(f_1 \circ \widetilde{f_2} \simeq id_{T_2}\) and a continuous \(\widetilde{f_3}: T_3 \to T_2\) such that \(\widetilde{f_3} \circ f_2 \simeq id_{T_2}\) and \(f_2 \circ \widetilde{f_3} \simeq id_{T_3}\); Step 2: see that \(\widetilde{f_2} \circ \widetilde{f_3} \circ f_2 \circ f_1 \simeq id_{T_1}\) and \(f_2 \circ f_1 \circ \widetilde{f_2} \circ \widetilde{f_3} \simeq id_{T_3}\).
Step 1:
There is a continuous \(\widetilde{f_2}: T_2 \to T_1\) such that \(\widetilde{f_2} \circ f_1 \simeq id_{T_1}\) and \(f_1 \circ \widetilde{f_2} \simeq id_{T_2}\), by Note for the definition of homotopy equivalence.
There is a continuous \(\widetilde{f_3}: T_3 \to T_2\) such that \(\widetilde{f_3} \circ f_2 \simeq id_{T_2}\) and \(f_2 \circ \widetilde{f_3} \simeq id_{T_3}\), as before.
Step 2:
Let us take \(\widetilde{f_2} \circ \widetilde{f_3}: T_3 \to T_1\).
\(\widetilde{f_2} \circ \widetilde{f_3}\) is continuous, by the proposition that for any maps between any arbitrary subspaces of any topological spaces continuous at any corresponding points, the composition is continuous at the point.
\(\widetilde{f_2} \circ \widetilde{f_3} \circ f_2 \circ f_1 = \widetilde{f_2} \circ (\widetilde{f_3} \circ f_2) \circ f_1 \simeq \widetilde{f_2} \circ id_{T_2} \circ f_1\), because \(\widetilde{f_3} \circ f_2 \simeq T_2\), by the proposition that for any homotopic maps from any 1st topological space into any 2nd topological space and any homotopic maps from the 2nd topological space into any 3rd topological space, the compositions of the homotopic maps are homotopic with a homotopy as this, \(= \widetilde{f_2} \circ f_1 \simeq id_{T_1}\).
\(f_2 \circ f_1 \circ \widetilde{f_2} \circ \widetilde{f_3} = f_2 \circ (f_1 \circ \widetilde{f_2}) \circ \widetilde{f_3} \simeq f_2 \circ id_{T_2} \circ \widetilde{f_3}\), because \(f_1 \circ \widetilde{f_2} \simeq id_{T_2}\), as before, \(= f_2 \circ \widetilde{f_3} \simeq id_{T_3}\).
So, \(f_2 \circ f_1\) is a homotopy equivalence.
