421: For 2 Path-Connected Points on Topological Space, There Is 'Groups - Group Homomorphisms' Isomorphism Between Fundamental Groups That Multiplies Inverse-Path Class from Left and Path Class from Right in Path Classes Groupoid

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A description/proof of that for 2 path-connected points on topological space, there is 'groups - group homomorphisms' isomorphism between fundamental groups that multiplies inverse-path class from left and path class from right in path classes groupoid

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Topics

About: topological space
About: group

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Note
• 3: Proof

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Starting Context

• The reader knows a definition of topological path-connected-ness of 2 points.
• The reader knows a definition of fundamental group.
• The reader knows a definition of %category name% isomorphism.
• The reader knows a definition of path classes groupoid.
• The reader admits the proposition that any groups map that maps the identity to the identity and maps any multiplication to the multiplication is a group homomorphism.
• The reader admits the proposition that any bijective group homomorphism is a 'groups - groups homomorphisms' isomorphism.

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Target Context

• The reader will have a description and a proof of the proposition that for any 2 path-connected points on any topological space, there is a 'groups - group homomorphisms' isomorphism between the fundamental groups with respect to the 2 points that (the isomorphism) multiplies the inverse-path class from the left and the path class from the right in the path classes groupoid.

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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 space, $T$, any path-connected points, $p,p^{\prime }\in T$, and any path, $\gamma :I\to T$, such that $\gamma (0)=p$ and $\gamma (1)=p^{\prime }$, there is a 'groups - group homomorphisms' isomorphism, $\varphi :{\pi }_1(T,p)\to {\pi }_1(T,p^{\prime })$, $[f]\mapsto [{\gamma }^{-1}][f][\gamma ]$ where $[{\gamma }^{-1}][f][\gamma ]$ is in the path classed groupoid.

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2: Note

$[{\gamma }^{-1}][f][\gamma ]$ makes sense because it is in the path classes groupoid: $[{\gamma }^{-1}]$, $[f]$, or $[\gamma ]$ is not any member of ${\pi }_1(T,p^{\prime })$ but is a member of the groupoid and the result is a member of ${\pi }_1(T,p^{\prime })$, because it is $[{\gamma }^{-1}\ast f\ast \gamma ]$, a loop that starts from $p^{\prime }$.

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3: Proof

$\varphi$ is injective, because for any $[f],[f^{\prime }]\in {\pi }_1(T,p)$, when $[{\gamma }^{-1}][f][\gamma ]=[{\gamma }^{-1}][f^{\prime }][\gamma ]$, $[f]=[e_p][f][e_p]=[\gamma ][{\gamma }^{-1}][f][\gamma ][{\gamma }^{-1}]=[\gamma ][{\gamma }^{-1}][f^{\prime }][\gamma ][{\gamma }^{-1}]=[e_p][f^{\prime }][e_p]=[f^{\prime }]$.

$\varphi$ is surjective, because for any $[f]\in {\pi }_1(T,p^{\prime })$, $[\gamma ][f][{\gamma }^{-1}]\in {\pi }_1(T,p)$, and $[{\gamma }^{-1}][\gamma ][f][{\gamma }^{-1}][\gamma ]=[e_{p^{\prime }}][f][e_{p^{\prime }}]=[f]$.

$\varphi$ is a group homomorphism, because $\varphi (e)=[{\gamma }^{-1}][e_p][\gamma ]=[{\gamma }^{-1}][\gamma ]=[e_{p^{\prime }}]=e$; $\varphi ([f][f^{\prime }])=[{\gamma }^{-1}][f][f^{\prime }][\gamma ]=[{\gamma }^{-1}][f][e_p][f^{\prime }][\gamma ]=[{\gamma }^{-1}][f][\gamma ][{\gamma }^{-1}][f^{\prime }][\gamma ]=\varphi ([f])\varphi ([f^{\prime }])$, by the proposition that any groups map that maps the identity to the identity and maps any multiplication to the multiplication is a group homomorphism.

So, $\varphi$ is a 'groups - group homomorphisms' isomorphism, by the proposition that any bijective group homomorphism is a 'groups - groups homomorphisms' isomorphism.

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References

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