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
Topics
About: topological space
About: group
The table of contents of this article
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.
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.
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: Description
For any topological space, \(T\), any path-connected points, \(p, p' \in T\), and any path, \(\gamma: I \to T\), such that \(\gamma (0) = p\) and \(\gamma (1) = p'\), there is a 'groups - group homomorphisms' isomorphism, \(\phi: \pi_1 (T, p) \to \pi_1 (T, p')\), \([f] \mapsto [\gamma^{-1}] [f] [\gamma]\) where \([\gamma^{-1}] [f] [\gamma]\) is in the path classed groupoid.
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 \(\pi_1 (T, p')\) but is a member of the groupoid and the result is a member of \(\pi_1 (T, p')\), because it is \([\gamma^{-1} * f * \gamma]\), a loop that starts from \(p'\).
3: Proof
\(\phi\) is injective, because for any \([f], [f'] \in \pi_1 (T, p)\), when \([\gamma^{-1}] [f] [\gamma] = [\gamma^{-1}] [f'] [\gamma]\), \([f] = [e_p] [f] [e_p] = [\gamma] [\gamma^{-1}] [f] [\gamma] [\gamma^{-1}] = [\gamma] [\gamma^{-1}] [f'] [\gamma] [\gamma^{-1}] = [e_p] [f'] [e_p] = [f']\).
\(\phi\) is surjective, because for any \([f] \in \pi_1 (T, p')\), \([\gamma] [f] [\gamma^{-1}] \in \pi_1 (T, p)\), and \([\gamma^{-1}] [\gamma] [f] [\gamma^{-1}] [\gamma] = [e_{p'}] [f] [e_{p'}] = [f]\).
\(\phi\) is a group homomorphism, because \(\phi (e) = [\gamma^{-1}] [e_p] [\gamma] = [\gamma^{-1}] [\gamma] = [e_{p'}] = e\); \(\phi ([f] [f']) = [\gamma^{-1}] [f] [f'] [\gamma] = [\gamma^{-1}] [f] [e_p] [f'] [\gamma] = [\gamma^{-1}] [f] [\gamma] [\gamma^{-1}] [f'] [\gamma] = \phi ([f]) \phi ([f'])\), 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, \(\phi\) is a 'groups - group homomorphisms' isomorphism, by the proposition that any bijective group homomorphism is a 'groups - groups homomorphisms' isomorphism.
