A description/proof of that canonical map from fundamental group on finite product topological space into product of constituent topological space fundamental groups is 'groups - group homomorphisms' isomorphism
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of product topology.
- The reader knows a definition of fundamental group.
- The reader knows a definition of fundamental group homomorphism induced by map.
- The reader knows a definition of product group.
- The reader knows a definition of %category name% isomorphism.
- 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 into each constituent topological space after the map is continuous.
- 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 - group homomorphisms' isomorphism.
Target Context
- The reader will have a description and a proof of the proposition that for any finite number of topological spaces and the product of the topological spaces, the canonical map from the fundamental group on the product topological space into the product of the fundamental groups on the constituent topological spaces is a 'groups - group homomorphisms' isomorphism.
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 finite number of topological spaces, \(T_1, T_2, ..., T_n\), the product topological space, \(T = T_1 \times T_2 \times ... \times T_n\), and any point, \(p = (p^1, p^2, ..., p^n) \in T\), the canonical map from the fundamental group on \(T\) into the product of the fundamental groups on \(T_i\)s, \(f: \pi_1 (T_1 \times T_2 \times ... \times T_n, (p^1, p^2, ..., p^n)) \to \pi_1 (T_1, p^1) \times \pi_1 (T_2, p^2) \times ... \times \pi_1 (T_n, p^n)\), \([f'] \mapsto (s_{1*} ([f']), s_{2*} ([f']), ..., s_{n*} ([f']))\), where \(f'\) is any continuous loop on \(T\) that starts at \(p\) and \(s_i: T_1 \times T_2 \times ... \times T_n \to T_i\) is the projection, is a 'groups - group homomorphisms' isomorphism.
2: Proof
\(s_{i*} ([f']) = [s_i \circ f']\) by the definition of fundamental group homomorphism induced by map. Product and inverse on any product group are component-wise product and component-wise inverse, respectively, by the definition of product group.
Let us prove that \(f\) is bijective. For any \([f']\) and \([f'']\) such that \([f'] \neq [f'']\) where \(f'\) and \(f''\) are continuous loops on \(T\) that starts at \(p = (p^1, p^2, ..., p^n) \in T\), let us suppose that \([s_i \circ f'] = [s_i \circ f'']\) for every \(i\). There would be a relative homotopy, \(F_i: I \times I \to T_i\), such that \((t, 0) \mapsto s_i \circ f' (t)\), \((t, 1) \mapsto s_i \circ f'' (t)\), and \(F_i (0, s) = s_i \circ f' (0) = s_i \circ f'' (0) = F_i (1, s) = s_i \circ f' (1) = s_i \circ f'' (1)\). Let us define \(F := I \times I \to T = (F_1, F_2, ..., F_n)\). \(F\) would be continuous, by the proposition that any map from any topological space to any product topological space is continuous if and only if the composition of the projection to each constituent topological space after the map is continuous. \(F (t, 0) = (s_1 \circ f' (t), s_2 \circ f' (t), ..., s_n \circ f' (t)) = f' (t)\), \(F (t, 1) = (s_1 \circ f'' (t), s_2 \circ f'' (t), ..., s_n \circ f'' (t)) = f'' (t)\), and \(F (0, s) = (s_1 \circ f' (0), s_2 \circ f' (0), ..., s_n \circ f' (0)) = f' (0) = (s_1 \circ f'' (0), s_2 \circ f'' (0), ..., s_n \circ f'' (0)) = f'' (0) = F (1, s) = (s_1 \circ f' (1), s_2 \circ f' (1), ..., s_n \circ f' (1)) = f' (1) = (s_1 \circ f'' (1), s_2 \circ f'' (1), ..., s_n \circ f'' (1)) = f'' (1)\). So, \([f'] = [f'']\), a contradiction. So, \(f\) is injective. For any \(([f'_1], [f'_2], ..., [f'_n]) \in \pi_1 (T_1, p^1) \times \pi_1 (T_2, p^2) \times ... \times \pi_1 (T_n, p^n)\), let us define \(f' := (f'_1, f'_2, ..., f'_n)\), a continuous loop on \(T\). Then, \(f ([f']) = (s_{1*} ([f']), s_{2*} ([f']), ..., s_{n*} ([f']))\), and \(s_{i*} ([f']) = [s_i \circ f'] = [f'_i]\). So, \(f\) is surjective. So, \(f\) is bijective.
Let us prove that \(f\) is a group homomorphism. \(s_{i*}\) is a group homomorphism, because it is a fundamental group homomorphism induced by map. \(f (e) = (s_{1*} (e), s_{2*} (e), ..., s_{n*} (e)) = (e, e, ..., e) = e\) where \(e\) is the identity elements. \(f ([f'] [f'']) = (s_{1*} ([f'] [f'']), s_{2*} ([f'] [f'']), ..., s_{n*} ([f'] [f''])) = (s_{1*} ([f']) s_{1*} ([f'']), s_{2*} ([f']) s_{2*} ([f'']), ..., s_{n*} ([f']) s_{n*} ([f''])) = (s_{1*} ([f']), s_{2*} ([f']), ..., s_{n*} ([f'])) (s_{1*} ([f'']), s_{2*} ([f'']), ..., s_{n*} ([f''])) = f ([f']) f ([f''])\). So, \(f\) is a group homomorphism, 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, \(f\) is a 'groups - group homomorphisms' isomorphism, by the proposition that any bijective group homomorphism is a 'groups - group homomorphisms' isomorphism.
