414: Compositions of Homotopic Maps Are Homotopic

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A description/proof of that compositions of homotopic maps are homotopic

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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

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Starting Context

• The reader knows a definition of homotopic map.
• The reader admits the proposition that any map from any topological space into any product topological space is continuous if and only if the composition of the projection of the map into each element of the product after the map is continuous.

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Target Context

• The reader will have a description and a proof of 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.

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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 topological spaces, $T_1,T_2,T_3$, any homotopic maps, $f_1,f_2:T_1\to T_2$, and any homotopic maps, $f_1^{\prime },f_2^{\prime }:T_2\to T_3$, $f_1^{\prime }\circ f_1$ and $f_2^{\prime }\circ f_2$ are homotopic.

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2: Proof

There are some homotopies, $f_3:T_1\times I\to T_2$ and $f_3^{\prime }:T_2\times I\to T_3$ such that $f_3(p,0)=f_1(p)$, $f_3(p,1)=f_2(p)$, $f_3^{\prime }(p,0)=f_1^{\prime }(p)$, and $f_3^{\prime }(p,1)=f_2^{\prime }(p)$.

Let us define $f^{″}:T_1\times I\to T_2\times I\to T_3$, $(p,i)\mapsto (f_3(p,i),i)\mapsto f_3^{\prime }(f_3(p,i),i)$. $f^{″}$ is continuous as a composition of continuous maps, while the 1st half of the composition is continuous by the proposition that any map from any topological space into any product topological space is continuous if and only if the composition of the projection of the map into each element of the product after the map is continuous. $f^{″}(p,0)=f_3^{\prime }(f_3(p,0),0)=f_3^{\prime }(f_1(p),0)=f_1^{\prime }(f_1(p))$. $f^{″}(p,1)=f_3^{\prime }(f_3(p,1),1)=f_3^{\prime }(f_2(p),1)=f_2^{\prime }(f_2(p))$.

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References

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